洛谷P2619 Tree I

经典的k条白边MST

带权二分，按照套路我们要选择尽量少的白边。

#include <cstdio>
#include <algorithm>

const int N = 100010;

int D;

struct Edge {
    int x, y, val, col;
    inline bool operator <(const Edge &w) const {
        if(val - D * col == w.val - D * w.col) {
            return col < w.col;
        }
        return val - D * col < w.val - D * w.col;
    }
}edge[N];

int n, m, k, ans;

namespace ufs{
    int fa[N];
    int find(int x) {
        if(x == fa[x]) {
            return x;
        }
        return fa[x] = find(fa[x]);
    }
    inline void merge(int x, int y) {
        fa[find(x)] = find(y);
        return;
    }
    inline bool check(int x, int y) {
        return find(x) == find(y);
    }
    inline void clear() {
        for(int i = 1; i <= n; i++) {
            fa[i] = i;
        }
        return;
    }
}

inline int check(int mid) {
    ufs::clear();
    D = mid;
    ans = 0;
    int cnt = 0;
    std::sort(edge + 1, edge + m + 1);
    for(int i = 1; i <= m; i++) {
        if(!ufs::check(edge[i].x, edge[i].y)) {
            cnt += edge[i].col;
            ans += edge[i].val - D * edge[i].col;
            ufs::merge(edge[i].x, edge[i].y);
        }
    }
    return cnt;
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= m; i++) {
        scanf("%d%d%d%d", &edge[i].x, &edge[i].y, &edge[i].val, &edge[i].col);
        edge[i].x++;
        edge[i].y++;
        edge[i].col ^= 1;
    }

    int l = -150, r = 150;
    while(l < r) {
        int mid = (l + r + 1) >> 1;
        int t = check(mid);
        //printf("[%d %d] mid = %d cnt = %d 
", l, r, mid, t);
        if(t == k) {
            printf("%d
", ans + k * mid);
            return 0;
        }
        if(t < k) {
            l = mid;
        }
        else {
            r = mid - 1;
        }
    }
    check(r);
    printf("%d
", ans + k * r);
    return 0;
}