HDU 1595find the longest of the shortest（spfa）

Problem Description

Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another.
Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed.
Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

 

Input

Each case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N.
In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.

 

Output

In the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.

 

Sample Input

  

   5 6
1 2 4
1 3 3
2 3 1
2 4 4
2 5 7
4 5 1

6 7
1 2 1
2 3 4
3 4 4
4 6 4
1 5 5
2 5 2
5 6 5

5 7
1 2 8
1 4 10
2 3 9
2 4 10
2 5 1
3 4 7
3 5 10
  

 

Sample Output

  

   11
13
27
  

 

Author

ailyanlu

 

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

 

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题目意思：先求1到n最短路，然后删除一条边，求得的第二长的1到n最短路的值？

思路：spfa的过程中记录路径，然后删除最短路中的边求得答案

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;

#define N 1005
#define INF 0x3f3f3f3f

int n,m,vis[N],use[N][N];
int head[N],dis[N];
int num,pre[N];

struct stud{
int to,next;
int va;
}e[N*N];

void build(int u,int v,int va)
{
	e[num].to=v;
	e[num].va=va;
	e[num].next=head[u];
	head[u]=num++;
}

void spfa()
{
	int i,j;
	queue<int>q;
	memset(dis,INF,sizeof(dis));
	memset(vis,0,sizeof(vis));
	vis[1]=1;
	dis[1]=0;
	q.push(1);
	pre[1]=-1;
	int cur;
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		vis[cur]=0;
		for(i=head[cur];i!=-1;i=e[i].next)
		{
			int x=e[i].to;
			if(dis[x]>dis[cur]+e[i].va)
			{
				dis[x]=dis[cur]+e[i].va;
				pre[x]=cur;
				if(!vis[x])
				{
					vis[x]=1;
					q.push(x);
				}
			}
		}
	}
}

int work()
{
	int i,j;
	memset(dis,INF,sizeof(dis));
	memset(vis,0,sizeof(vis));
	queue<int>q;
	q.push(1);
	vis[1]=1;
	dis[1]=0;
	int cur;

	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		vis[cur]=0;
		for(i=head[cur];i!=-1;i=e[i].next)
		{
			int x=e[i].to;
			if(use[cur][x]&&dis[x]>dis[cur]+e[i].va)
			{
				dis[x]=dis[cur]+e[i].va;
				if(!vis[x])
				{
					q.push(x);
					vis[x]=0;
				}
			}
		}
	}
	return dis[n];
}

int main()
{
	int i,j,u,v,va;
	while(~scanf("%d%d",&n,&m))
	{
		memset(use,0,sizeof(use));
		memset(head,-1,sizeof(head));
		num=0;
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&va);
			use[u][v]=use[v][u]=1;
			build(u,v,va);
			build(v,u,va);
		}
		spfa();
		int ans=dis[n];

		for(i=n;pre[i]!=-1;i=pre[i])
		{
			int x=pre[i];
			use[i][x]=use[x][i]=0;
			ans=max(work(),ans);
			use[i][x]=use[x][i]=1;
		}
		printf("%d\n",ans);
	}
	return 0;
}