HDU-3665（单源最短路）

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

---

思路：

在Dijkstra的基础上稍作修改即可，注意0x7fffffff可能会超范围

---

#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x7ffffff
using namespace std;

int N;
int G[][];
int tmp;
int sea[];
int a,b;
int s[];
int d[];
int minn;
int v;
int ans;
int st[];

int min(int a,int b)
{
    return a<b?a:b;
}

int main()
{
    while(~scanf("%d",&N))
    {
        if(N == ){
            printf("0\n");
            continue;
        }
        ans = INF;
        memset(s,,sizeof(s));
        for(int i = ;i < N;i++)
            for(int j = ;j < N;j++)
                G[i][j] = i==j?:INF;
        for(int i = ;i < N;i++) {
            scanf("%d%d",&tmp,&sea[i]);
            while(tmp--) {
                scanf("%d%d",&a,&b);
                G[i][a] = b;
            }
        }
        if(sea[]){
            printf("0\n");
            continue;
        }
        s[] = ;
        for(int i = ;i < N;i++)
            d[i] = G[][i];
        for(int i = ;i < N;i++)
        {
            minn = INF;
            for(int j = ;j < N;j++)
                if(!s[j] && d[j]<minn) minn = d[v=j];
            s[v] = ;
            if(sea[v]) ans = min(ans,minn);
            for(int j = ;j < N;j++)
                if(!s[j] && d[j]>G[v][j]+minn) d[j] = G[v][j]+minn;
        }
        printf("%d\n",ans);
    }
    return ;
}