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Problem Description
The final battle is coming. Now Harry Potter is located at city 1, and Voldemort is located at city n. To make the world peace as soon as possible, Of course, Harry Potter will choose the shortest road between city 1 and city n. But unfortunately, Voldemort is so powerful that he can choose to destroy any one of the existing roads as he wish, but he can only destroy one. Now given the roads between cities, you are to give the shortest time that Harry Potter can reach city n and begin the battle in the worst case.
Input
First line, case number t (t<=20).
Then for each case: an integer n (2<=n<=1000) means the number of city in the magical world, the cities are numbered from 1 to n. Then an integer m means the roads in the magical world, m (0< m <=50000). Following m lines, each line with three integer u, v, w (u != v,1 <=u, v<=n, 1<=w <1000), separated by a single space. It means there is a bidirectional road between u and v with the cost of time w. There may be multiple roads between two cities.
Output
Each case per line: the shortest time to reach city n in the worst case. If it is impossible to reach city n in the worst case, output “-1”.
Sample Input
3
4
4
1 2 5
2 4 10
1 3 3
3 4 8
3
2
1 2 5
2 3 10
2
2
1 2 1
1 2 2
Sample Output
15
-1
2
Author
tender@WHU
Source
2011 Multi-University Training Contest 15 - Host by WHU
思路：由于数据范围还行，考虑枚举最短路上的每条边，把这条边删除后跑一遍spfa。
#include<bits/stdc++.h>using namespace std;const int maxn = 5e4+5;const int inf=1e9+100;int d[maxn];struct cxk{ int v,w,id;};int flag=1,p[maxn],pre[maxn],num[maxn];vector<cxk>g[maxn];bool vis[maxn];void spfa(int x){ memset(vis,false,sizeof(vis)); for(int i=0;i<maxn;++i) d[i]=inf; d[x]=0; queue<int>q; q.push(x); vis[x]=true; while(!q.empty()) { int top=q.front(); q.pop(); vis[top]=false; for(auto v:g[top]) { if(p[v.id]) continue; if(d[top]+v.w<d[v.v]) { d[v.v]=d[top]+v.w; if(flag) { pre[v.v]=top;num[v.v]=v.id; } if(!vis[v.v]) q.push(v.v),vis[v.v]=true; } } }}int main(){ int T,n,u,v,w,m; scanf("%d",&T); while(T--) { memset(p,0,sizeof(p)); memset(pre,0,sizeof(pre)); memset(num,0,sizeof(num)); flag=1; scanf("%d %d",&n,&m); for(int i=0;i<=n;++i) g[i].clear(); for(int i=1;i<=m;++i) { scanf("%d %d %d",&u,&v,&w); u--,v--; g[u].push_back({ v,w,i}); g[v].push_back({ u,w,i}); } spfa(0); if(d[n-1]==inf) { printf("-1\n");continue; } flag=0; int ans=0,flag1=0; for(int i=n-1;i!=0;i=pre[i]) { p[num[i]]=1; spfa(0); p[num[i]]=0; if(d[n-1]==inf) { flag1=1;printf("-1\n");break; } ans=max(ans,d[n-1]); } if(flag1) continue; if(ans==inf) puts("-1"); else printf("%d\n",ans); }}转载地址：http://hrewz.baihongyu.com/