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原數據出錯，現已更新 by liutian,但未重測---2016.6.24

Source

對於圖上的最小生成樹

如果我們得到的最小生成樹上的白邊小於$need$條，那麽說明白邊的權值整體偏大，

那麽我們考慮對所有的白邊減去一個權值，這樣最小生成樹上的白邊就會變多

這個過程很顯然具有單調性，於是可以二分減去的權值

註意一個坑：當權值相同的時候優先選擇白邊

#include<cstdio>
#include
 
<algorithm>
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1; 
    while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * f;
}
 
int N, M, need;
struct Edge {
    int u, v, w, opt;
    bool operator <(const Edge &rhs) const {
        return w == rhs.w ? opt < rhs.opt : w < rhs.w;
    }
}E[MAXN], e[MAXN];
int Val = 0, fa[MAXN];
int siz[MAXN];
int find(int x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
int unionn(int x, int y) {
    int fx = find(x), fy = find(y);
    if(siz[fx] < siz[fy]) swap(fx, fy);
    fa[fy] = fx;
}
bool check(int x) {
    Val = 0;
    for(int i = 1; i <= M; i++) {
        E[i] = e[i];
        if(e[i].opt == 0) E[i].w += x;
    }
    for(int i = 1; i <= N; i++) 
        fa[i] = i, siz[i] = 1;
    int tot = 0, white = 0;
    sort(E + 1, E + M + 1);
    for(int i = 1; i <= M; i++) {
        if(find(E[i].u) != find(E[i].v)) {
            unionn(E[i].u, E[i].v);
            Val += E[i].w; tot++;
            if(E[i].opt == 0) white++;
        }
        if(tot == N - 1) break;
    }
    return white >= need;
}
main() {
    #ifdef WIN32
    freopen("a.in", "r", stdin);
    #endif
    N = read(), M = read(), need = read();
    for(int i = 1; i <= M; i++) {
        int x = read() + 1, y = read() + 1, z = read(), opt = read();
        e[i] = (Edge){x, y, z, opt};
    }
    int l = -110, r = 110, ans = 0;
    while(l <= r) {
        int mid = l + r >> 1;
        if(check(mid)) ans = Val - mid * need, l = mid + 1;
        else r = mid - 1;    
    } 
    printf("%d", ans);
}

BZOJ 2654: tree(二分 最小生成樹)