127.0.0.1:9200
URL: http://127.0.0.1:9200/likecs_art_db/_search
REQUEST:
Array
(
    [query] => Array
        (
            [match] => Array
                (
                    [text] => Array
                        (
                            [query] => Codeforces Round #485 (Div. 2)
                        )

                )

        )

    [highlight] => Array
        (
            [fields] => Array
                (
                    [text] => stdClass Object
                        (
                        )

                )

            [pre_tags] => #em#
            [post_tags] => #/em#
        )

    [size] => 8
    [from] => 0
)
RESPONSE:
string(7237) "{"took":12,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":48.516438,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"19861","_score":48.516438,"_source":{"id":"19861","text":"Codeforces Round #485 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"fzl194","tagsname":"","tagsid":"[]","catesname":"Codeforces","catesid":"[\"1494\"]","createtime":"1527681070"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# ##em#485#/em# (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"74370","_score":36.49176,"_source":{"id":"74370","text":"Codeforces Round #585 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"31415926535x","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1568905097"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #585 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"87297","_score":36.49176,"_source":{"id":"87297","text":"Codeforces Round #604 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"SwiftAC","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1577331742"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #604 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"92357","_score":36.49176,"_source":{"id":"92357","text":"Codeforces Round #618 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"hezongdnf","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1581301910"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #618 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"92438","_score":36.49176,"_source":{"id":"92438","text":"Codeforces Round #618 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"StarRoadTang","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1581346141"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #618 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"104795","_score":36.49176,"_source":{"id":"104795","text":"Codeforces Round #643 (Div.2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"Anonytt","tagsname":"Codeforces","tagsid":"[\"12974\"]","catesname":"","catesid":"[]","createtime":"1589690332"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #643 (#em#Div#/em#.#em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"103348","_score":36.49176,"_source":{"id":"103348","text":"Codeforces Round #639 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"et3-tsy","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1588839128"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #639 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"114716","_score":36.49176,"_source":{"id":"114716","text":"Codeforces Round #673 (Div. 2) ","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"william4s","tagsname":"","tagsid":"[]","catesname":"codeforces\u9898\u89e3","catesid":"[9605]","createtime":"1601280449"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #673 (#em#Div#/em#. #em#2#/em#)"]}}]}}"
127.0.0.1:9200
URL: http://127.0.0.1:9200/likecs_art_db/_search
REQUEST:
Array
(
    [query] => Array
        (
            [match] => Array
                (
                    [text] => Array
                        (
                            [query] => Codeforces Round #485 (Div. 2)
                        )

                )

        )

    [highlight] => Array
        (
            [fields] => Array
                (
                    [text] => stdClass Object
                        (
                        )

                )

            [pre_tags] => #em#
            [post_tags] => #/em#
        )

    [size] => 8
    [from] => 8
)
RESPONSE:
string(7301) "{"took":11,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":48.516438,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"117427","_score":36.49176,"_source":{"id":"117427","text":"Codeforces Round #678 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"blacktion","tagsname":"","tagsid":"[]","catesname":"Codeforces","catesid":"[\"1494\"]","createtime":"1603599102"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #678 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"18403","_score":36.49176,"_source":{"id":"18403","text":"Codeforces Round #483 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"fzl194","tagsname":"","tagsid":"[]","catesname":"Codeforces","catesid":"[\"1494\"]","createtime":"1526565964"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #483 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"49014","_score":36.49176,"_source":{"id":"49014","text":"Codeforces Round #533 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"HQHQ","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1548051154"},"highlight":{"text":["#em#Codeforces#/em# #em#Round#/em# #533 (#em#Div#/em#. #em#2#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"122728","_score":36.49176,"_source":{"id":"122728","text":"Codeforces Round #695 (Div. 2)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        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Codeforces Round #485 (Div. 2) - 爱码网
fzl194

第2次CF,前半个小时发现前面3题都是水题,直接1A,第4题是求最短路,用BFS跑了一下,感觉题目的限制应该不会超时,交了以后也过了,第5题是关于随机排列的题目,模拟了交换最少次数,交了之后也过了,第6题是求图的连通分量,但是图很大,边是由两个数字的&性质决定的,用一个set,学会了边删除边遍历。由于一直在queue中,睡一觉后发现还是没过,但是值得庆幸的是我的其他5题没有被hack。所以排到了134名,感觉妥妥的,rateing飙升160,现在有1660蓝名,希望接下来到暑假可以冲击紫名。

 

http://codeforces.com/contest/987

A:太水略过

B:求x^y 和y^x大小

取对数,比赛的时候怕精度问题,用了eps判等于,结果过了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 int main()
 5 {
 6     double x, y, eps = 1e-12;
 7     cin >> x >> y;
 8     double ans = x * log(y) - y * log(x);
 9     if(fabs(ans) < eps)cout<<"="<<endl;
10     else if(ans > 0) cout<<"<"<<endl;
11     else cout<<">"<<endl;
12     return 0;
13 }

 

C:给数组s和c,满足i<j<k 且 si < sj < sk 使ci + cj + ck最小

直接DP,DP[i][j]为以i结尾长度为j的最大价值。最终求最小的DP[i][3]

初始化DP[i][1] = s[i]

注意,dp[i][2]中i从2开始递推,dp[i][3]中i从3开始递推,为了防止出错,写了两个二重循环。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 5000;
 5 const int INF = 1e9 + 7;
 6 ll a[maxn], b[maxn], dp[maxn][5];
 7 int main()
 8 {
 9     int n;
10     cin >> n;
11     for(int i = 1; i <= n; i++)cin >> a[i];
12     for(int i = 1; i <= n; i++)cin >> b[i], dp[i][1] = b[i];
13     for(int i = 2; i <= 3; i++)
14         for(int j = 1; j <= n; j++)dp[j][i] = INF;
15     ll ans = INF;
16     for(int i = 2; i <= n; i++)
17     {
18         for(int j = 1; j < i; j++)
19             if(a[j] < a[i])
20             dp[i][2] = min(dp[i][2], dp[j][1] + b[i]);
21     }
22     for(int i = 3; i <= n; i++)
23     {
24         for(int j = 2; j < i; j++)
25             if(a[j] < a[i])
26             dp[i][3] = min(dp[i][3], dp[j][2] + b[i]);
27     }
28     for(int i = 1; i <= n; i++)ans = min(ans, dp[i][3]);
29     if(ans == INF)cout<<"-1"<<endl;
30     else cout<<ans<<endl;
31     return 0;
32 }

 

D:n个城市,m条无向路,共有k种货物,每个城市只有一种货物,让你求在城市i举办交易会,至少有s种货物参加,求最少的路径之和

由于s和k最大为100,可以用BFS跑每一个点,跑到货物种类等于s的时候返回路径和(这里是种类不是数量)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 100000 + 10;
 5 const int INF = 1e9 + 7;
 6 bool vis[maxn], v[maxn];
 7 int dis[maxn], a[maxn];
 8 vector<int>G[maxn];
 9 int n, m, k, S;
10 int BFS(int s)
11 {
12     memset(vis, 0,sizeof(vis));
13     memset(v, 0,sizeof(v));
14     vis[s] = 1;
15     dis[s] = 0;
16     int tot = 0, now, ans = 0, u;
17     queue<int>q;
18     q.push(s);
19     while(!q.empty())
20     {
21         now = q.front();
22         q.pop();
23         if(!v[a[now]])
24         {
25             v[a[now]] = 1;
26             tot++;
27             ans += dis[now];
28             if(tot == S)return ans;
29         }
30         for(int i = 0; i < G[now].size(); i++)
31         {
32             u = G[now][i];
33             if(!vis[u])
34             {
35                 vis[u] = 1;
36                 dis[u] = dis[now] + 1;
37                 q.push(u);
38             }
39         }
40     }
41 }
42 int main()
43 {
44     scanf("%d%d%d%d", &n, &m, &k, &S);
45     for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
46     int u, v;
47     while(m--)
48     {
49         scanf("%d%d", &u, &v);
50         G[u].push_back(v);
51         G[v].push_back(u);
52     }
53     //BFS(1);
54     printf("%d", BFS(1));
55     for(int i = 2; i <= n; i++)printf(" %d", BFS(i));
56     puts("");
57     return 0;
58 }

 

 

E:给出一个1-n随机排列,Petr的随机化排列方法是随机交换3*n次,Um_nik的随机化排列方法是随机交换7*n+1次。问给出的排列是谁随机化得到的。

初始排列 1,2,3...n

模拟了一下最小交换次数tot,如果3*n -  tot是偶数的话,那么就是Petr的,否则就是Um_nik的

数组a存储序列,数组b的b[i]表示a[j] = i的j的值

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1000000 + 10;
 5 const int INF = 1e9 + 7;
 6 int a[maxn], b[maxn];
 7 int main()
 8 {
 9     int n;
10     cin >> n;
11     for(int i = 1; i <= n; i++)
12     {
13         scanf("%d", &a[i]);
14         b[a[i]] = i;
15     }
16     int tot = 0, u;
17     for(int i = 1; i <= n; i++)
18     {
19         if(a[i] == i)continue;
20         tot++;
21         u = b[i];
22         swap(b[i], b[a[i]]);
23         swap(a[i], a[u]);
24     }
25     tot = 3 * n - tot;
26     if(tot % 2 == 0)cout<<"Petr"<<endl;
27     else cout<<"Um_nik"<<endl;
28     return 0;
29 }

 

F:给n和m,还有m个互不相同的数,均小于1<<n,如果m个数字中a & b = 0,那么a 和b有边相连,问有多少个连通分量

比赛的时候,用set存储每个数,每次取出第一个数,在set中删去与他相邻的点,重复,但是还是超时了,不过学到了如何用set边删除边遍历。

对于迭代器it,如果需要删除it指向的位置

  先用it2 = it存下这个位置

  it++

  s.erase(it2)

这样就可以遍历的时候删除了

题目正解应该是,把每个数看做一个01集合,对于在m个数中的数,求它的补集的子集个数即可

这里用4位来举个例子,

比如5(0101),补集:(1010),补集的子集:(1010)  (1000) (0010) (0000)

补集的子集和该集合的&运算的结果为0,满足条件。

那么就可以枚举0 - ((1<<n) - 1)进行DFS即可。每次找到所有联通的点。

时间复杂度O(1<<n)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = (1<<23);
 5 bool vis[maxn], num[maxn];
 6 int n, m;
 7 void dfs(int x)
 8 {
 9     if(vis[x])return;
10     vis[x] = 1;
11     for(int i = 0; i < n; i++)//遍历x的子集
12     {
13         if(x & (1 << i))
14             dfs(x ^ (1 << i));
15     }
16     if(num[x])dfs(((1<<n) - 1) & (~x));//如果是num中的元素,就寻找x的补集的子集
17 }
18 int main()
19 {
20     int x;
21     scanf("%d%d", &n, &m);
22     for(int i = 1; i <= m; i++)
23     {
24         scanf("%d", &x);
25         num[x] = 1;
26     }
27     int ans = 0;
28     for(int i = 0; i < (1 << n); i++)
29     {
30         if(num[i] && !vis[i])
31             dfs(((1<<n) - 1) & (~i)), ans++;//dfs其补集
32     }
33     cout<<ans<<endl;
34     return 0;
35 }

 

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