bzoj2654: tree(最小生成树+二分)

题目

bzoj2654: tree

解析

kruscal在做最小生成树时先按权值排序，权值小的先被选到，我们可以通过控制白色边的边权来控制白色边的数量。
我们可以通过二分答案来给白边加某一个值
同时注意两点

1. 不要忘记减去给白边加的值
2. 排序时白边优先

代码

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

int n, m, num, need, ans, tot, sum, cnt;

int fa[N], head[N], w[N];

struct node {
    int u, v, w, col;
    bool operator <(const node &oth) const {
        return w == oth.w ? col < oth.col : w < oth.w;
    }
} e[N], tmp[N];

int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int kruscal() {
    cnt = 0, sum = 0, ans = 0;
    for (int i = 1; i <= n; ++i) fa[i] = i;
    sort(e + 1, e + 1 + m);
    for (int i = 1; i <= m; ++i) {
        int x = find(e[i].u), y = find(e[i].v);
        if (x == y) continue;
        fa[x] = y;
        ans += e[i].w;
        if (!e[i].col) sum++;
        cnt++;
        if (cnt == n - 1) break;
    }
    return sum;
}

bool check(int x) {
    for (int i = 1; i <= m; ++i) {
        e[i] = tmp[i];
        if (!e[i].col) e[i].w += x;
    }
    return kruscal() >= need;
}

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> need;
    for (int i = 1, x, y, z, c; i <= m; ++i) {
        cin >> x >> y >> z >> c;
        e[i] = (node) {x + 1, y + 1, z, c};
        tmp[i] = (node) {x + 1, y + 1, z, c};
    }
    int l = -1000, r = 1000;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) l = mid + 1, tot = ans - need * mid;
        else r = mid - 1;
    }
    cout << tot;
    return 0;
}

bzoj2654: tree(最小生成树+二分)

原文：https://www.cnblogs.com/lykkk/p/11223011.html