题目连接：
http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=129723
题意：
给你一个n点m图的边
1到n有多条最短路，问你所有经过的边的总和*2是多少
题解：
对1,n分别求单源最短路径上spfa
枚举某条边是否为最短上的边
即 边权+disA[i] + disB[i] = 最短路长度，就是答案
#include<bits/stdc++.h> using namespace std; const int N = 1e6+20, M = 1e6+10, mod = 1e9+7, inf = 1e9+1000; typedef long long ll;int head[N],t,vis[N],dis[N],n,m,disA[N],disB[N]; struct edge{int to,value,next;}e[N*2];void add(int u,int v,int w) {e[t].next=head[u],e[t].to=v,e[t].value=w;head[u]=t++;}void init(){t = 0;memset(head,-1,sizeof(head)); }void spfa(int u) {for(int i=1;i<=n;i++) dis[i] = inf, vis[i] = 0;vis[1] = 1;dis[u] = 0;queue<int> q;q.push(u);while(!q.empty()) {int k = q.front();q.pop();vis[k] = 0;for(int i=head[k];i!=-1;i=e[i].next) {int to = e[i].to;if(dis[k] + e[i].value < dis[to]) {dis[to] = dis[k] + e[i].value;if(!vis[to]) {q.push(to);vis[to] = 1;}}}} } int main() {while(~scanf("%d%d",&n,&m)) {init();for(int i=1;i<=m;i++) {int a,b,c;scanf("%d%d%d",&a,&b,&c);a++,b++;add(a,b,c);add(b,a,c);}spfa(1);for(int i=1;i<=n;i++) disA[i] = dis[i];spfa(n);for(int i=1;i<=n;i++) disB[i] = dis[i];int mx = dis[1];ll ans = 0;for(int i=1;i<=n;i++) {for(int j=head[i];j!=-1;j=e[j].next) {int to = e[j].to;int value = e[j].value;if(disA[i] + disB[to] + value == mx) ans+=value;}}printf("%lld\n",2ll*ans);} }
