标签：并查集

find the most comfortable road

4 4
1 2 2
2 3 4
1 4 1
3 4 2
2
1 3
1 2

1
0

分析：先按照速度从小到大排序，然后从当前i开始到n建立并查集，直到初始和目的形成连接，然后与ans比较

代码：

/*hdoj 1598*/ 
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define M 1005
#define INF 0x3f3f3f3f
using namespace std;

struct node{
    int from, to, sp;
}s[M];
int fat[M], n, m;

int f(int x){
    if(fat[x] != x) fat[x] = f(fat[x]);
    return fat[x];
}

int cmp(node a, node b){
    return a.sp<b.sp;    
}

int main(){
    int i, a, b, c, j, q;
    while(scanf("%d%d", &n, &m) == 2){
        for(i = 0; i < m; i ++)
            scanf("%d%d%d", &s[i].from, &s[i].to, &s[i].sp);
        sort(s, s+m, cmp);
        
        scanf("%d", &q);
        while(q --){
            int ans = INF;
            scanf("%d%d", &a, &b);
            for(i = 0; i < m; i ++){
                for(j = 1; j <= n; j ++) fat[j] = j;
                for(j = i; j < m; j ++){
                    int x = f(s[j].from);
                    int y = f(s[j].to);
                    if(x!=y) fat[x] = y;
                    if(f(a) == f(b)){
                        ans = min(ans, s[j].sp-s[i].sp); break;
                    }
                }
            }
            if(ans == INF) printf("-1\n");
            else printf("%d\n", ans);
        }
    }
    return 0;
}

hdoj 1598 find the most comfortable road 【并查集】+【暴力枚举】

标签：并查集

原文地址：http://blog.csdn.net/shengweisong/article/details/41056839