HDU 1598 Find the most comfortable road 二分枚举答案 或 并查集

我是用的二分，这样似乎要慢点

并查集的那个没写。。

窝真的好菜。。

#include <vector>
#include <queue>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <iostream>
using namespace std;
const int MAX = 205;
const int MAXM = 1005;
int spd[MAXM];
bool vis[MAX];

struct _edge {
    int to, spd;
};
vector<_edge> mp[MAX];

void init(int n){
    for(int i = 0; i <= n; ++i)
        mp[i].clear();
}

bool bfs(int s, int t, int l, int h) {
    memset(vis, false, sizeof(vis));
    vis[s] = true;
    queue<int> tag;
    tag.push(s);
    int sz, to, tp;
    while(!tag.empty()){
        tp = tag.front();
        tag.pop();
        sz = mp[tp].size();
        for(int i = 0; i < sz; ++i){
            _edge& e = mp[tp][i];
            if(vis[e.to] || e.spd < l || e.spd > h)
                continue;
            if(e.to == t)
                return true;
            vis[e.to] = true;
            tag.push(e.to);
        }
    }
    return false;
}

int main() {
    int n, m, q, s, t, sp;
    while(~scanf("%d%d", &n, &m)){
        int mx = 0;
        init(n);
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d", &s, &t, &sp);
            mp[s].push_back((_edge){t, sp});
            mp[t].push_back((_edge){s, sp});
            spd[i] = sp;
            mx = max(mx, sp);
        }
        sort(spd, spd + m);
        scanf("%d", &q);
        while(q --){
            scanf("%d%d", &s, &t);
            int l = 0, r = mx, mid, ans = -1;
            bool ok = false;
            while(l <= r){
                mid = (l + r) >> 1;
                ok = false;
                for(int i = 0; spd[i] + mid < mx; ++i){
                    if(bfs(s, t, spd[i], spd[i] + mid)){
                        ok = true;
                        break;
                    }
                }
                if(ok){
                    ans = mid;
                    r = mid - 1;
                }
                else
                    l = mid + 1;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}