HDU 3665 Seaside (Floyd 最短路)

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1146    Accepted Submission(s): 834

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

51 01 12 02 33 11 14 1000 10 1

 

Sample Output

2

 

Source

2010 Asia Regional Harbin

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3665

题目大意：有n个镇，编号为0~n-1，有些镇靠海，xiaoY现在在0号镇，求出从0号镇到海边的最短路径长度。

题目分析：根据题意建下图，Floyd跑一下，再找个最短的

#include <cstdio>int const INF = 0xffffff;int n, map[15][15];int ok[15];void Floyd()  {      for(int i = 0; i < n; i++)          for(int j = 0; j < n; j++)              for(int k = 0; k < n; k++)                  if(map[j][i] + map[i][k] < map[j][k])                      map[j][k] = map[j][i] + map[i][k];  } void Init()  {      for(int i = 0; i < n; i++)          for(int j = 0; j < n; j++)              i == j ? map[i][j] = 0 : map[i][j] = INF;  }  int main(){    while(scanf("%d", &n) != EOF)    {        Init();        int m, s, l;        for(int i = 0; i < n; i++)        {            scanf("%d %d", &m, &ok[i]);            for(int j = 0; j < m; j++)            {                scanf("%d %d", &s, &l);                map[i][s] = l;            }        }        Floyd();        int ans = INF;        for(int i = 0; i < n; i++)            if(ok[i] && ans > map[0][i])                ans = map[0][i];        printf("%d\n", ans);    }}