F - Silver Cow Party POJ - 3268

最新推荐文章于 2021-08-25 21:46:55 发布

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#poj #最短路径

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本文介绍了一种求解从多个起点到指定终点再返回起点的最短路径问题的方法，并通过一个具体的例子展示了如何使用最短路径算法来找到所有奶牛参加聚会并返回所需的最大时间。

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

最短路径+思维

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=110000+100;
const int inf=0x3f3f3f3f;
int cnt=0;
int n,m,x;
int head[maxn],vis[maxn],dis[maxn],diss[maxn];
int xx[maxn],yy[maxn],zz[maxn];
struct edge
{
	int v,next,val;
}edge[maxn*3+1];
struct node
{
	int point,distance;
	node(){}
	node(int _point,int _distance)
	{
		point=_point;
		distance=_distance;
	}
	friend bool operator < (node aa,node bb)
	{
		return aa.distance>bb.distance;
	}
};
void add_edge(int u,int v,int w)
{
	edge[cnt].v=v;
	edge[cnt].val=w;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void zyz(int s)
{
	int i;
	for(i=0;i<=n;i++)
	{
		dis[i]=inf;
	} 
	dis[s]=0;
	priority_queue<node> q;
	q.push(node(s,dis[s]));
	while(!q.empty())
	{
		node now;
		now=q.top();
		q.pop();
		if(vis[now.point])
			continue;
		//printf("11111\n");
		for(i=head[now.point];i!=-1;i=edge[i].next)
		{
			int to=edge[i].v;
			//printf("%d %d %d\n",to,dis[now.point],edge[i].val);
			if(dis[to]>dis[now.point]+edge[i].val)
			{
				dis[to]=dis[now.point]+edge[i].val;
				//printf("%d %d\n",to,dis[to]);
				q.push(node(to,dis[to]));
			}
		}
	}
}
int main ()
{
	while(~scanf("%d%d%d",&n,&m,&x))
	{
		int i;
		cnt=0;
		memset(vis,0,sizeof(vis));
		memset(head,-1,sizeof(head));
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&xx[i],&yy[i],&zz[i]);
			add_edge(xx[i],yy[i],zz[i]);
		}
		zyz(x);
		memset(diss,0,sizeof(diss));
		for(i=1;i<=n;i++)
		{
			diss[i]=dis[i];
			//printf("%d %d\n",i,dis[i]);
		}
		memset(vis,0,sizeof(vis));
		memset(head,-1,sizeof(head));
		cnt=0;
		for(i=1;i<=m;i++)
		{
			add_edge(yy[i],xx[i],zz[i]);
		}
		zyz(x);	
		//for(i=1;i<=n;i++)
			//printf("%d %d\n",i,dis[i]);	
		int maxx=0;
		for(i=1;i<=m;i++)
		{
			if(dis[i]!=inf&&diss[i]!=inf)
			{
				dis[i]=dis[i]+diss[i];
				maxx=max(maxx,dis[i]);
			}
		}
		printf("%d\n",maxx);
	}
}