poj3268（单源最短路）

Silver Cow Party

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 23064 Accepted: 10550

Description

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 T_i (1 ≤ T_i ≤ 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?

Input

Line 1: Three space-separated integers, respectively: N, M, and X 
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

10

三种最短路算法复杂度一比较就知道应该选用dijkstra算法，而由于边是单向的，所以返回路径要单独算，再算一次dijkstra，这是这次把边的起点终点换一下，解决。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;#define INF 1e9+1#define MAX_E 100005#define MAX_V 1005int cost[MAX_V][MAX_V];int d[MAX_V];int d2[MAX_V];bool used[MAX_V];int V,M,X;void init(){for(int i=0;i<MAX_V;i++){for(int j=0;j<MAX_V;j++){cost[i][j]=INF;}cost[i][i]=0;}}void dijkstra(int s){fill(d,d+V,INF);fill(used,used+V,false);d[s]=0;while(1){int v=-1;for(int u=0;u<V;u++){if(!used[u]&&(v==-1||d[u]<d[v]))v=u;}if(v==-1)break;used[v]=true;for(int u=0;u<V;u++){d[u]=min(d[u],d[v]+cost[v][u]);}}}void Fake_dijkstra(int s){fill(d2,d2+V,INF);fill(used,used+V,false);d2[s]=0;while(1){int v=-1;for(int u=0;u<V;u++){if(!used[u]&&(v==-1||d2[u]<d2[v]))v=u;}if(v==-1)break;used[v]=true;for(int u=0;u<V;u++){d2[u]=min(d2[u],d2[v]+cost[u][v]);      //将边的起点与终点交换}}}int main(){init();cin>>V>>M>>X;int a,b,t;for(int i=0;i<M;i++){cin>>a>>b>>t;cost[a-1][b-1]=t;}dijkstra(X-1);Fake_dijkstra(X-1);int maxt=0;for(int i=0;i<V;i++)if(d[i]+d2[i]>maxt)maxt=d[i]+d2[i];cout<<maxt<<endl;}