hdu 1595刪邊加最短路
find the longest of the shortest
Time Limit: 1000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4862 Accepted Submission(s): 1819
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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卡了將近兩個小時,注意只標記刪邊之前
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1005
#define inf 0x3f3f3f3f
int dis[maxn][maxn];
int vis[maxn];
int step[maxn];
int d[maxn];
int flag;
int n,m;
int dij(int s)
{memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
d[i]=i==s?0:inf;
for(int i=1;i<=n;i++)
{
int x,minn=inf;
for(int y=1;y<=n;y++)
if(!vis[y]&&minn>=d[y])
minn=d[x=y];
vis[x]=1;
for(int y=1;y<=n;y++)
if(d[y]>d[x]+dis[x][y])
{d[y]=d[x]+dis[x][y];
if(!flag)
step[y]=x;
}
}
return d[n];
}
int main()
{
while(~scanf("%d%d",&n,&m))
{int i,j;
for(i=1; i<=n; i++)
{
step[i]=0;
for(j=1; j<=n; j++)
{
if(i!=j)
dis[i][j]=inf;
else
dis[i][j]=0;
}
}
int a,b,c;
for(i=0; i<m; i++)
{
scanf("%d%d%d",&a,&b,&c);
if(dis[a][b]>c)
dis[a][b]=dis[b][a]=c;
}
flag=0;
int ans=dij(1);
flag=1;
for(i=n; i!=1; i=step[i])
{
int tt=dis[i][step[i]];
dis[i][step[i]]=dis[step[i]][i]=inf;
int ans1=dij(1);///從0開始不再記錄路徑
ans=max(ans,ans1);
dis[i][step[i]]=dis[step[i]][i]=tt;
}
printf("%d\n",ans);
}
return 0;
}