HDU 3665 Seaside

题目链接：

http://acm.split.hdu.edu.cn/showproblem.php?pid=3665

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

 

Hint:

题意：

一个含有N个小城镇的大城镇，以0到N-1的顺序，先给出两个数a,b;表示i到其他城镇有a条道路,b表示该地方是否有海，0表示没有海，1表示有海。

然后在给出a1,x;a1表示i到a1之间有道路，距离是x。求从0到有海的城市的最小的距离。

题解：

简单的最短路问题。

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10+10;
#define met(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
int map[maxn][maxn];
int visited[maxn],dis[maxn];
int num[maxn];
int n,m;
void dijkstra(int x)
{
    int min=0,p=0;
    for(int i=0;i<n;i++)
    {
        dis[i]=map[x][i];
        visited[i]=0;
    }
    visited[x]=1;
    for(int i=0;i<n;i++)
    {
        min=inf;
        for(int j=0;j<n;j++)
        {
            if(!visited[j]&&dis[j]<min)
                {
                    min=dis[j];
                    p=j;
                }
        }
        visited[p]=1;
        for(int j=0;j<n;j++)
        {
            if(!visited[j]&&dis[p]+map[p][j]<dis[j])
                dis[j]=dis[p]+map[p][j];
        }
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            map[i][j]=inf;
        met(num,0);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&m,&num[i]);
            for(int j=0;j<m;j++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                map[i][x]=map[x][i]=y;
            }
        }
        dijkstra(0);
        int ans=inf;
        for(int i=0;i<n;i++)
        {
            if(num[i]==1)
                ans=min(ans,dis[i]);
        }
        printf("%d
",ans);
    }
}