杭电 3665 Seaside（最短路）（弗洛伊德过）

最新推荐文章于 2018-07-27 15:25:08 发布

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在智能视频分析中的重要作用，包括其在自动驾驶、AR增强现实等场景中的实际应用。

无脑模板题~

Seaside

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1452 Accepted Submission(s): 1055

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

Sample Output

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int panduan[121];
long long int dis[121][121];
int n;
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                dis[i][j]=0x1f1f1f1f;
            }
        }
        memset(panduan,0,sizeof(panduan));
        for(int i=0;i<n;i++)
        {
            int a,b,c,d;
            scanf("%d%d",&a,&b);
            panduan[i]=b;
            for(int j=0;j<a;j++)
            {
                scanf("%d%d",&c,&d);
                dis[i][c]=dis[c][i]=d;
            }
        }
        int i,j,k;
            for(k=0;k<n;k++)
                for(i=0;i<n;i++)
                    for(j=0;j<n;j++)
                        if(dis[i][j]>dis[i][k]+dis[k][j])
                           dis[i][j]=dis[i][k]+dis[k][j];
        int output=0x1f1f1f1f;
        for(int i=0;i<n;i++)
        {
            if(panduan[i])
            {
                if(dis[0][i]<output)output=dis[0][i];
            }
        }
        printf("%d\n",output);
    }
}