hdoj 3665 Seaside【最短路】

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1356    Accepted Submission(s):
973

Problem Description

XiaoY is living in a big city, there are N towns in it
and some towns near the sea. All these towns are numbered from 0 to N-1 and
XiaoY lives in the town numbered ’0’. There are some directed roads connecting
them. It is guaranteed that you can reach any town from the town numbered ’0’,
but not all towns connect to each other by roads directly, and there is no ring
in this city. One day, XiaoY want to go to the seaside, he asks you to help him
find out the shortest way.

 

Input

There are several test cases. In each cases the first
line contains an integer N (0<=N<=10), indicating the number of the towns.
Then followed N blocks of data, in block-i there are two integers, Mi
(0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads
beginning with the i-th town. Pi indicates whether the i-th town is near to the
sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two
integers S_Mi and L_Mi, which means that the distance
between the i-th town and the S_Mi town is L_Mi.

 

Output

Each case takes one line, print the shortest length
that XiaoY reach seaside.

 

Sample Input

5

1 0

1 1

2 0

2 3

3 1

1 1

4 100

0 1

0 1

 

Sample Output

2

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f
#define MAX 110
#define minn(x,y)(x<y?x:y)
using namespace std;
int map[MAX][MAX];
int vis[MAX],used[MAX],low[MAX];
int n,m;
void init()
{
    int i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            map[i][j]=i==j?0:INF;
}
void solve()
{
    int i,j,min,next,mindis=0;
    memset(vis,0,sizeof(vis));
    for(i=0;i<n;i++)
        low[i]=map[0][i];
    vis[0]=1;
    for(i=1;i<n;i++)
    {
        min=INF;
        for(j=0;j<n;j++)
        {
            if(!vis[j]&&min>low[j])
            {
                next=j;
                min=low[j];
            }
        }
        vis[next]=1;
        for(j=0;j<n;j++)
        {
            if(!vis[j]&&low[j]>map[next][j]+low[next])
                low[j]=map[next][j]+low[next];
        }
    }
    int sum=INF;
    for(i=0;i<n;i++)
    {
        if(used[i]==1)
        sum=minn(sum,low[i]);
    }
    printf("%d\n",sum);
}
int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        init();
        int a,b,c;
        int road,sea;
        memset(used,0,sizeof(used));
        int ok=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&road,&sea);
            if(sea==1)
                used[i]=1;
            while(road--)
            {
                scanf("%d%d",&a,&b);
                if(map[i][a]>b)
                    map[i][a]=b;
            }
        }
        if(used[0]==1)
        {
            printf("0\n");
            ok=1;
            continue;
        }
        if(!ok)
        solve();
    }
    return 0;
}