Heavy Transportation POJ - 1797 SPFA 最短路

最新推荐文章于 2021-09-24 11:14:23 发布

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本文介绍了一个寻找从起点到终点的最大可运输重量的问题解决方案。通过使用SPFA算法进行路径搜索，并针对每条路径上的最小允许重量进行调整，可以找出在给定城市地图中，从Hugo的起点到顾客目的地的最大运输能力。

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Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

Sample Output

Scenario #1:

求1到n点的所有通路中最小值的最大情况。用SPFA，改变一下判断条件。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<string.h>
#define inf 0x3f3f3f3f 
using namespace std;
const int maxn=1005;
int n,m,t,tot;
int head[maxn],dis[maxn];
bool vis[maxn];
struct node
{
	int to;
	int w;
	int next;
}e[maxn*maxn];

void init()
{
	memset(head,-1,sizeof(head));
	memset(vis,false,sizeof(vis));
	memset(dis,0,sizeof(dis));
	tot=0;
	return ;
}
void addedge(int u,int v,int w)
{
	e[tot].to=v;
	e[tot].w=w;
	e[tot].next=head[u];
	head[u]=tot++;
	return ;
}
void spfa(int st)
{
	queue<int>q;
	int v,w;
	vis[st]=true;
	dis[st]=inf;
	q.push(st);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=false;
		for(int i=head[u];i!=-1;i=e[i].next)
		{
			v=e[i].to;
			w=e[i].w;
			if(min(dis[u],w)>dis[v])
			{
				dis[v]=min(dis[u],w);
				if(!vis[v])
				{
					q.push(v);
					vis[v]=true;
				}
			}
			
		}
	}
}
int main() 
{
	//freopen("C:/input.txt", "r", stdin);
	scanf("%d",&t);
	for(int i=1;i<=t;i++)
	{
		init();
		scanf("%d%d",&n,&m);
		for(int i=0;i<m;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			addedge(u,v,w);
			addedge(v,u,w);
		}
		spfa(1);
		printf("Scenario #%d:\n",i);
		printf("%d\n",dis[n]);
		if(i!=t)
		{
			printf("\n");
		}
	}
    return 0;
}