HDU 5636 关键点的 floyd 最短路问题

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 285 Accepted Submission(s): 92

Problem Description

There is a path graph

.

You are given the graph and several queries about the shortest path between some pairs of vertices.

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test cases, output an integer

-th query.

Sample Input

1 10 2 2 4 5 7 8 10 1 5 3 1

Sample Output

Source

BestCoder Round #74 (div.2)

bc 题对于我们这种渣渣来说太感人了

看了题解做的比赛的时候传统的最短路都会超时不用想想dfs 但姿势表达太渣

求最短路初始时单链每点间距离为1 另添加距离为1的3条边也就是由关键的6个点

关键6个点floyd 处理具体看代码理解这个都好久还是太弱

在询问中 6*6 种与查询区间枚举比较找到最短路

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<vector>
 5 #include<cmath>
 6 #define ll __int64
 7 using namespace std;
 8 ll t;
 9 ll n,q;
10 ll a[10];
11 ll dis[10][10];
12 int main()
13 {
14     scanf("%I64d",&t);
15     for(int i=1;i<=t;i++)
16     {
17         memset(dis,0,sizeof(dis));
18         scanf("%I64d %I64d",&n,&q);
19         scanf("%I64d %I64d %I64d %I64d %I64d %I64d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]);
20         ll sum=0;
21         for(int j=1;j<=6;j++)
22                 for(int k=1;k<=6;k++)
23             {
24                 dis[j][k]=abs(a[j]-a[k]);
25             }
26         dis[1][2]=1;
27         dis[2][1]=1;
28         dis[3][4]=1;
29         dis[4][3]=1;
30         dis[5][6]=1;
31         dis[6][5]=1;
32         for(int i=1;i<=6;i++)
33          for(int k=1;k<=6;k++)
34              for(int m=1;m<=6;m++)
35         {
36             if(dis[k][i]+dis[i][m]<dis[k][m])
37                 dis[k][m]=dis[k][i]+dis[i][m];
38         }
39         for(int j=1;j<=q;j++)
40         {
41             ll aa,bb,t;
42             scanf("%I64d %I64d",&aa,&bb);
43             if(aa>bb)
44             {
45                 t=aa;
46                 aa=bb;
47                 bb=t;
48             }
49             ll ans=abs(bb-aa);
50             for(int k=1;k<=6;k++)
51                 for(int m=1;m<=6;m++)
52             {
53                 if(ans>abs(aa-a[k])+abs(bb-a[m])+dis[k][m])
54                     ans=abs(aa-a[k])+abs(bb-a[m])+dis[k][m];
55             }
56             sum=(sum+j*ans)%1000000007;
57         }
58      printf("%I64d\n",sum);
59     }
60     return 0;
61 }

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