E - Easy Dijkstra Problem（求最短路）

Description

Determine the shortest path between the specified vertices in the graph given in the input data.
Hint: You can use Dijkstra's algorithm.
Hint 2: if you're a lazy C++ programmer, you can use set and cin/cout (with sync_with_stdio(0)) - it should suffice.

Input

first line - one integer - number of test cases For each test case the numbers V, K (number of vertices, number of edges) are given,Then K lines follow, each containing the following numbers separated by a single space:a_i, b_i, c_i ,It means that the graph being described contains an edge from a_i to b_i,with a weight of c_i.Below the graph description a line containing a pair of integers A, B is present.The goal is to find the shortest path from vertex A to vertex B.All numbers in the input data are integers in the range 0..10000.

Output

For each test case your program should output (in a separate line) a single number C - the length of the shortest path from vertex A to vertex B. In case there is no such path, your program should output a single word "NO" (without quotes)

Example

Input:
3
3 2
1 2 5
2 3 7
1 3
3 3
1 2 4
1 3 7
2 3 1
1 3
3 1
1 2 4
1 3

Output:
12
5
NO
解题思路：坑题，WA了好几发=_=||原来题目说明的是顶点a到顶点b是一条有向边，即a-->b，而不是无向图求单源最短路，裸题（邻接矩阵）水过！
AC代码：

 #include<iostream>
 #include<string.h>
 #include<cstdio>
 using namespace std;
 const int INF=0x3f3f3f3f;
 const int maxn=;
 int t,n,k,a,b,c,st,ed,dis[maxn],cost[maxn][maxn];bool flag,vis[maxn];
 void dijkstra(){
     for(int i=;i<=n;++i)
         dis[i]=cost[st][i];
     dis[st]=;vis[st]=true;
     for(int i=;i<n;++i){
         int k=-;
         for(int j=;j<=n;++j)
             if(!vis[j]&&(k==-||dis[k]>dis[j]))k=j;
         if(dis[k]==INF){flag=true;break;}//如果此时没有最小值即为INF，说明肯定是达不到终点ed，直接退出循环
         if(k==-)break;
         vis[k]=true;
         for(int j=;j<=n;++j)
             if(!vis[j])dis[j]=min(dis[j],dis[k]+cost[k][j]);
     }
 }
 int main(){
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&k);
         for(int i=;i<=n;++i)
             for(int j=;j<=n;++j)
                 cost[i][j]=cost[j][i]=(i==j?:INF);
         memset(vis,false,sizeof(vis));
         while(k--){
             scanf("%d%d%d",&a,&b,&c);
             cost[a][b]=min(cost[a][b],c);//去重
         }
         scanf("%d%d",&st,&ed);
         flag=false;
         dijkstra();
         if(flag)printf("NO\n");
         else printf("%d\n",dis[ed]);
     }
     return ;
 }