Silver Cow Party（好题）

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 ofM (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; roadi requiresT_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, andX
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers:A_i,B_i, andT_i. The described road runs from farmA_i to farmB_i, requiringT_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.

题意：有n个农场，编号1~N，农场里奶牛将去X号农场。这N个农场之间有M条单向路（注意）,通过第i条路将需要花费Ti单位时间。选择最短时间的最优路径来回一趟,花费在去的路上和返回农场的这些最优路径的最长时间是多少？

思路：计算出每头牛去X并且回来的最短路径所需要的时间，然后求出这n-1个农场的牛的最长时间即可，两次运用dijkstra；

1 计算回来的时间：以X为源点，求出源点到各个农场的最短路径；

2 计算去的时间：将路径反转，在用一次dij，求源点到农场的最短路径（实则求牛去X的最短路径）；

3 取两次和求最大值；

本题类似于 次小生成树----hpu王小二，都是同样的程序跑了两次，异曲同工吧；

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int graph[1010][1010],lowcost[1010],low[1010],mst[1010];int x,n,m;void dij(int v)//dijkstra模板；{int i,j,min,minid;memset(mst,0,sizeof(mst));mst[v]=1;for(i=1;i<=n;i++)lowcost[i]=graph[v][i];for(i=2;i<=n;i++){min=INF;for(j=1;j<=n;j++){if(!mst[j] && lowcost[j]<min){min=lowcost[j];minid=j;}}mst[minid]=1;for(j=1;j<=n;j++){if(!mst[j] && lowcost[j]>graph[minid][j]+lowcost[minid])lowcost[j]=graph[minid][j]+lowcost[minid];} }}int main(){int i,j,time,wa;while(scanf("%d %d %d",&n,&m,&x)!=EOF){memset(graph,INF,sizeof(graph));while(m--){scanf("%d %d %d",&i,&j,&time);if(time<graph[i][j])graph[i][j]=time;//单向路径； }dij(x);//计算牛回去的最短时间； for(i=1;i<=n;i++)//执行完dij,lowcost[i]代表以x为起点,i为终点的最短路径;low[i]=lowcost[i];for(i=1;i<=n;i++){for(j=i+1;j<=n;j++)//j不能从1开始,否则反转两次,相当于没变化； {wa=graph[j][i];graph[j][i]=graph[i][j];graph[i][j]=wa;}}dij(x);int maxn=0,ans;for(i=1;i<=n;i++){if(i!=x)//此条件不可少； ans=low[i]+lowcost[i];//来回时间相加； maxn=max(maxn,ans);}printf("%d\n",maxn);}return 0;}

bellman-ford，思路一样，只是换种方式，和dijkstra 时间差不多；

#include<stdio.h>#include<string.h>#define INF 0x3f3f3f3fint mst[300000],mst1[300000];int n,m,x;struct node{int a;int b;int c;}s[300000];void bellman_ford(int v){memset(mst,INF,sizeof(mst));mst[v]=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(mst[s[j].a]>s[j].c+mst[s[j].b])//单向路径；mst[s[j].a]=s[j].c+mst[s[j].b];}}}int main(){int a,b,c,wa;while(scanf("%d %d %d",&n,&m,&x)!=EOF){for(int i=1;i<=m;i++)scanf("%d %d %d",&s[i].a,&s[i].b,&s[i].c);bellman_ford(x);for(int i=1;i<=n;i++)//去的最短路径；mst1[i]=mst[i];for(int i=1;i<=m;i++)//置换一下；{wa=s[i].a;s[i].a=s[i].b;s[i].b=wa;}bellman_ford(x);//计算回来的最短路径；int ans,maxn=0;for(int i=1;i<=n;i++){if(i!=x)ans=mst[i]+mst1[i];if(ans>maxn) maxn=ans;}printf("%d\n",maxn);}return 0;}