One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
10
是很典型的最短路径问题，只不过这里是来回的，路径的单向的，为了不超时用了效率较高的SPFA，先求x点到每点的最短路径（以x点为起点SPFA一次），然后求每个点到x点的最短路径（以每个点为起点SPFA一次），然后相加找最大
代码：
#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
#define M 102
int n,m,x;
int ans[1005],dis[1005],team[2050],num[1005],a[1005][1005],w[1005][1005];
bool vis[1005];
int main()
{int i,j,k,u,s,e,t;scanf("%d%d%d",&n,&m,&x);memset(w,0x7f,sizeof(w));for(i=1;i<=m;i++){scanf("%d%d%d",&s,&e,&t);a[s][num[s]++]=e;w[s][e]=t;}memset(dis,0x7f,sizeof(dis));dis[x]=0;memset(vis,false,sizeof(vis));team[1]=x; vis[x]=true;int head=0,tail=1;do{head++;head=((head-1)%2050)+1;u=team[head];vis[u]=false;for(j=0;j<num[u];j++)if(dis[a[u][j]]>dis[u]+w[u][a[u][j]]){dis[a[u][j]]=dis[u]+w[u][a[u][j]];if(!vis[a[u][j]]){tail++;head=((head-1)%2050)+1;team[tail]=a[u][j];vis[a[u][j]]=true;}}}while(head!=tail);for(i=1;i<=n;i++)ans[i]=dis[i];for(i=1;i<=n;i++){if(i==x) continue;memset(team,0,sizeof(team));memset(dis,0x7f,sizeof(dis));dis[i]=0;memset(vis,false,sizeof(vis));team[1]=i; vis[i]=true;int head=0,tail=1;do{head++;head=((head-1)%2050)+1;u=team[head];vis[u]=false;for(j=0;j<num[u];j++)if(dis[a[u][j]]>dis[u]+w[u][a[u][j]]){dis[a[u][j]]=dis[u]+w[u][a[u][j]];if(!vis[a[u][j]]){tail++;head=((head-1)%2050)+1;team[tail]=a[u][j];vis[a[u][j]]=true;}}}while(head!=tail);ans[i]+=dis[x];}int maxn=0;for(i=1;i<=n;i++)if(ans[i]>maxn) maxn=ans[i];printf("%d\n",maxn);return 0;
}