POJ 3268-Silver Cow Party【正向+反向最短路】

Silver Cow Party

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 17121 Accepted: 7815

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; roadi 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 farmA_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

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

Source

USACO 2007 February Silver

解题思路：

题目大意就是有n头牛m条路，x最为目的地问在这个有向途中，回去再回来，最短路的情况下的花费，最小话费取厥于，最大的哪一个，将x这位出发点。正向和反向建图。

记住是有向图

#include<stdio.h>#include<cstring>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int mapz[1100][1100];int mapf[1100][1100];int disz[1100],disf[1100];bool vis[1100];int n,m,x;void PP1(int x){int i,j;for(i=1;i<=n;i++){disz[i]=mapz[x][i];}vis[x]=true;disz[x]=0;for(i=0;i<n-1;i++){int pos=x,min=INF;for(j=1;j<=n;j++){if(!vis[j]&&disz[j]<min){pos=j;min=disz[j];}}vis[pos]=true;for(j=1;j<=n;j++){if(!vis[j]&&disz[j]>disz[pos]+mapz[pos][j]){disz[j]=disz[pos]+mapz[pos][j];}}}}void PP2(int x){int i,j;for(i=1;i<=n;i++){disf[i]=mapf[x][i];}vis[x]=true;disf[x]=0;for(i=0;i<n-1;i++){int pos=x,min=INF;for(j=1;j<=n;j++){if(!vis[j]&&disf[j]<min){pos=j;min=disf[j];}}vis[pos]=true;for(j=1;j<=n;j++){if(!vis[j]&&disf[j]>disf[pos]+mapf[pos][j]){disf[j]=disf[pos]+mapf[pos][j];}}}}int main(){while(scanf("%d%d%d",&n,&m,&x)!=EOF){int i,j;memset(mapz,INF,sizeof(mapz));memset(mapf,INF,sizeof(mapf));for(i=0;i<m;i++){int xx,yy,tt;scanf("%d%d%d",&xx,&yy,&tt);mapz[xx][yy]=min(mapz[xx][yy],tt);mapf[yy][xx]=min(mapf[yy][xx],tt);}memset(vis,false,sizeof(vis));PP1(x);memset(vis,false,sizeof(vis));PP2(x);int ans=-1;for(i=1;i<=n;i++){ans=max(ans,disz[i]+disf[i]);}printf("%d\n",ans);}return 0;}