POJ.3268 Silver Cow Party (Dijkstra)-优快云博客

本文详细解析了POJ.3268 Silver Cow Party问题的Dijkstra算法实现，通过两个方向的最短路径计算，找出所有牛从聚会返回所需的最大时间。代码中展示了初始化矩阵、遍历调整、Dijkstra算法的具体实现。

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POJ.3268 Silver Cow Party (Dijkstra)

题意分析

稍后补

代码总览

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#define nmax 1005
#define inf 1e8+7
using namespace std;
int n,m,x;
int mp[nmax][nmax];
int shorttime[2][nmax];
bool visit[nmax];
void dij(int s,int hh)
{
    int cur  = s;
    for(int i = 1;i<=n;++i){
        shorttime[hh][i]= inf;
        visit[i] = false;
    }
    shorttime[hh][cur] = 0;
    visit[cur] = true;
    for(int i =2;i<=n;++i){
        int nowmin = inf;
        for(int j = 1;j<=n;++j){
            if(!visit[j] && shorttime[hh][cur] + mp[cur][j] < shorttime[hh][j]){
                shorttime[hh][j] = shorttime[hh][cur] + mp[cur][j];
            }
        }
        for(int j = 1;j<=n;++j){
            if(!visit[j] && nowmin >shorttime[hh][j] ){
                cur = j;
                nowmin = shorttime[hh][j];
            }
        }
        visit[cur] = true;
    }
}
void init()
{
    for(int i =1 ;i<=n;++i){
        for(int j = 1;j<=n;++j){
            mp[i][j] = inf;
        }
    }
    for(int i = 1;i<=m;++i){
        int a,b,d;
        scanf("%d %d %d",&a,&b,&d);
        mp[a][b] = d;
    }
}
void traverse()
{
    for(int i = 1;i<=n;++i){
        for(int j = 1;j<=i;++j){
            swap(mp[i][j],mp[j][i]);
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d %d %d",&n,&m,&x) != EOF){
        init();
        int ans[nmax]= {0};
        int maxt = 0;
        dij(x,0);
        traverse();
        dij(x,1);
        for(int i = 1;i<=n;++i){
            if(i == x) continue;
            ans[i] = shorttime[0][i] + shorttime[1][i];
            if(ans[i]> maxt){
                maxt = ans[i];
            }

        }
        printf("%d\n",maxt);
    }
    return 0;
}