BZOJ 1196 二分答案+并查集

http://www.lydsy.com/JudgeOnline/problem.php?id=1196

题目大意：n个城市，m-1条路，每条路有一级公路和二级公路之分，你要造n-1条路，一级公路至少要造k条，求出所造路的最大所需的val的最小值.

思路：首先我们一定要明确这个不是一题求所有花费的最小值的问题。然后我们只要二分答案就可以了。最后注意一下条件的拜访即可。

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
const int maxn = 10000 + 5;
const int inf = 0x3f3f3f3f;
struct Edge{
    int u, v, val1, val2;
    Edge(int u = 0, int v = 0, int v1 = 0, int v2 = 0): u(u), v(v), val1(v1), val2(v2){}
    bool operator < (const Edge &a) const{
        if (val1 != a.val1) return val1 < a.val1;
        return val2 < a.val2;
    }
}e[maxn * 2];
int n, k, m;
int par[maxn];
int pfind(int x){
    if (par[x] == x) return x;
    return par[x] = pfind(par[x]);
}

bool judge(int midval){
    for (int i = 1; i <= n; i++) par[i] = i;
    int cnt1 = 0, cnt = 0;
    for (int i = 1; i <= m - 1; i++){
        Edge a = e[i];
        int pu = pfind(a.u), pv = pfind(a.v);
        if (pu == pv) continue;
        if (a.val1 <= midval) cnt1++, cnt++, par[pu] = pv;
        else if (a.val2 <= midval) cnt++, par[pu] = pv;
    }
    if (cnt == n-1 && cnt1 >= k) return true;
    return false;
}

int main(){
    scanf("%d%d%d", &n, &k, &m);
    int lb = 0, rb = 30000;
    for (int i = 1; i <= m - 1; i++){
        int u, v, v1, v2;
        scanf("%d%d%d%d", &u, &v, &v1, &v2);
        e[i] = Edge(u, v, v1, v2);
    }
    sort(e + 1, e + m);
    while (lb < rb){
        int mid = lb + (rb - lb) / 2;
        if (judge(mid)){
            rb = mid;
        }
        else {
            lb = mid + 1;
        }
    }
    printf("%d\n", lb);
    return 0;
}
/*
3 1 3
1 2 11 4
2 3 10 2
*/