【POJ3268】Silver Cow Party 最短路

标签：poj3268   最短路   

题意：一堆奶牛去某个地方，去了又回，然后求去回和的最大值。

题解：两遍最短路，结束，邻接矩阵存边可以避免建反图。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N  1005
#define inf 0x3f3f3f3f
using namespace std;

int map[N][N],n,m,s;
int dist1[N],dist2[N];
bool visit[N];
void dij1()
{
	memset(dist1,0x3f,sizeof(dist1));
	memset(visit,0,sizeof(visit));
	int i,u,v,temp,round;
	dist1[s]=0;
	for(round=1;round<n;round++)
	{
		temp=inf;
		for(i=1;i<=n;i++)
		{
			if(!visit[i]&&temp>dist1[i])
			{
				temp=dist1[i];
				u=i;
			}
		}
		visit[u]=1;
		for(v=1;v<=n;v++)
		{
			dist1[v]=min(dist1[v],dist1[u]+map[u][v]);
		}
	}
	return ;
}
void dij2()
{
	memset(dist2,0x3f,sizeof(dist2));
	memset(visit,0,sizeof(visit));
	int i,u,v,temp,round;
	dist2[s]=0;
	for(round=1;round<n;round++)
	{
		temp=inf;
		for(i=1;i<=n;i++)
		{
			if(!visit[i]&&temp>dist2[i])
			{
				temp=dist2[i];
				u=i;
			}
		}
		visit[u]=1;
		for(v=1;v<=n;v++)
		{
			dist2[v]=min(dist2[v],dist2[u]+map[v][u]);
		}
	}
	return ;
}

int main()
{
//	freopen("test.in","r",stdin);
	int i,a,b,c;
	scanf("%d%d%d",&n,&m,&s);
	memset(map,0x3f,sizeof(map));
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		map[a][b]=c;
	}
	dij1();
	dij2();
	int ans=0;
	for(i=1;i<=n;i++)ans=max(ans,dist1[i]+dist2[i]);
	printf("%d\n",ans);
	return 0;
}

【POJ3268】Silver Cow Party 最短路

标签：poj3268   最短路   

原文地址：http://blog.csdn.net/vmurder/article/details/40736745