题目

链接

http://poj.org/problem?id=3268

字面描述

Silver Cow Party
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 46808 Accepted: 20917
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 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?

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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output

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

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
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

思路

这道题有来回求，但并没有直接给出源点。
开始我们已x为源点，向其他点做最短路
接下来任意一点均要到x点做最短路
最后比较出两者之和的最大值

代码实现

#include
#include
#include
#include
using namespace std;

const int maxn=1e3+10;
const int maxm=1e5+10;
int n,m,x,cnt,ans;
int head[maxn],dis[maxn],dis1[maxn],dis2[maxn];
bool vis[maxn];
struct node{
	int v,w,nxt;
}e[maxm];
inline bool add(int u,int v,int w){
	e[++cnt].v=v;
	e[cnt].w=w;
	e[cnt].nxt=head[u];
	head[u]=cnt;
}
inline void Dijkstra(int u){
	queue<int>q;
	memset(vis,false,sizeof(vis));
	memset(dis,0x3f,sizeof(dis));
	dis[u]=0;
	vis[u]=true;
	q.push(u);
	while(!q.empty()){
		int x=q.front();
		q.pop();
		vis[x]=false;
		for(int i=head[x];i;i=e[i].nxt){
			if(dis[e[i].v]>dis[x]+e[i].w){
				dis[e[i].v]=dis[x]+e[i].w;
				if(!vis[e[i].v]){
					vis[e[i].v]=true;
					q.push(e[i].v);
				}
			}
		}
	}
} 
int main(){
	scanf("%d%d%d",&n,&m,&x);
	for(int i=1;i<=m;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);
	}
	Dijkstra(x);
	for(int i=1;i<=n;i++)dis1[i]=dis[i];
	for(int i=1;i<=n;i++){
		Dijkstra(i);
		dis2[i]=dis[x];
		ans=max(ans,dis1[i]+dis2[i]);
	} 
	printf("%d\n",ans);
	return 0;
}
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