2018 Multi-University Training Contest 5 Solution

A - Always Online

Unsolved.

 

B - Beautiful Now

Solved.

题意：

给出一个n, k  每次可以将n这个数字上的某两位交换，最多交换k次，求交换后的最大和最小值

思路：

很明显有一种思路，对于最小值，尽可能把小的放前面， 对于最大值，尽可能把打的放前面。但是如果有多个最小数字或者最大数字会无法得出放哪个好，因此BFS一下即可

  1 #include<bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 const int INF = 0x3f3f3f3f;
  6 const int maxn = 1e5 + 10;
  7 
  8 struct node{
  9     int num, idx, step;
 10     node(){}
 11     node(int num, int idx, int step) :num(num), idx(idx), step(step){}
 12 };
 13 
 14 int n, k;
 15 int cnt;
 16 int arr[maxn];
 17 
 18 int BFS1()
 19 {
 20     queue<node>q;
 21     q.push(node(n, cnt, 0));
 22     int ans = INF;
 23     while(!q.empty())
 24     {
 25         node now = q.front();
 26         q.pop();
 27         ans = min(ans, now.num);
 28         if(now.step == k) continue;
 29         if(now.idx == 1) continue;
 30         int tmp = now.num;
 31         cnt = 0;
 32         while(tmp)
 33         {
 34             arr[++cnt] = tmp % 10;
 35             tmp /= 10;
 36         }
 37         int Min = now.idx;
 38         for(int i = now.idx - 1; i >= 1; --i)
 39         {
 40             if(arr[i] == arr[now.idx]) continue;
 41             if(arr[i] == 0 && now.idx == cnt) continue;
 42             if(arr[i] < arr[Min]) Min = i;
 43         }
 44         if(Min == now.idx)
 45         {
 46             q.push(node(now.num, now.idx - 1, now.step));
 47         }
 48         else
 49         {
 50             for(int i = now.idx - 1; i >= 1; --i)
 51             {
 52                 if(arr[i] == arr[Min])
 53                 {
 54                     swap(arr[i], arr[now.idx]);
 55                     tmp = 0;
 56                     for(int j = cnt; j >= 1; --j)
 57                     {
 58                         tmp = tmp * 10 + arr[j];
 59                     }
 60                     q.push(node(tmp, now.idx - 1, now.step + 1));
 61                     swap(arr[i], arr[now.idx]);
 62                 }
 63             }
 64         }
 65     }
 66     return ans;
 67 }
 68 
 69 int BFS2()
 70 {
 71     queue<node>q;
 72     q.push(node(n, cnt, 0));
 73     int ans = -INF;
 74     while(!q.empty())
 75     {
 76         node now = q.front();
 77         q.pop();
 78         ans = max(ans, now.num);
 79         if(now.step == k) continue;
 80         if(now.idx == 1) continue;
 81         int tmp = now.num;
 82         cnt = 0;
 83         while(tmp)
 84         {
 85             arr[++cnt] = tmp % 10;
 86             tmp /= 10;
 87         }
 88         int Max = now.idx;
 89         for(int i = now.idx - 1; i >= 1; --i)
 90         {
 91             if(arr[i] == arr[now.idx]) continue;
 92             if(arr[i] > arr[Max]) Max = i;
 93         }
 94         if(Max == now.idx)
 95         {
 96             q.push(node(now.num, now.idx - 1, now.step));
 97         }
 98         else
 99         {
100             for(int i = now.idx - 1; i >= 1; --i)
101             {
102                 if(arr[i] == arr[Max])
103                 {
104                     swap(arr[i], arr[now.idx]);
105                     tmp = 0;
106                     for(int j = cnt; j >= 1; --j)
107                     {
108                         tmp = tmp * 10 + arr[j];
109                     }
110                     q.push(node(tmp, now.idx - 1, now.step + 1));
111                     swap(arr[i], arr[now.idx]);
112                 }
113             }
114         }
115     }
116     return ans;
117 }
118 
119 int main()
120 {
121     int t;
122     scanf("%d", &t);
123     while(t--)
124     {
125         scanf("%d %d", &n ,&k);
126         int tmp = n;
127         cnt = 0;
128         while(tmp)
129         {
130             cnt++;
131             tmp /= 10;
132         }
133         k = min(k, cnt - 1);
134         int ans1 = BFS1();
135         int ans2 = BFS2();
136         printf("%d %d\n", ans1, ans2);
137     }
138     return 0;
139 }

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