【BZOJ2654】tree【二分】【最小生成树】

来源：互联网发布：2017淘宝卖什么成本低编辑：程序博客网时间：2024/06/04 08:06

【题目链接】

奇怪的二分。

考虑给白边的边权加上一个数，这个数越大，MST时选的白边就越少。

注意排序时候，如果边权相等，要先选白边。

/* Forgive me Not */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 50005, maxm = 100005, inf = 0x3f3f3f3f;int n, m, k, fa[maxn];struct edge {int u, v, w, c;bool operator < (const edge &x) const {return w != x.w ? w < x.w : c < x.c;}} g[maxm], e[maxm];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);}inline int check(int mid) {int cnt = 0, res = 0;for(int i = 1; i <= n; i++) fa[i] = i;for(int i = 1; i <= m; i++) {e[i] = g[i];if(!e[i].c) e[i].w += mid;}sort(e + 1, e + 1 + m);for(int i = 1; i <= m; i++) {int u = find(e[i].u), v = find(e[i].v);if(u != v) {res += e[i].w;if(!e[i].c) cnt++;fa[u] = v;}}return cnt >= k ? res : -1;}int main() {n = iread(); m = iread(); k = iread();for(int i = 1; i <= m; i++) {int u = iread(), v = iread(), w = iread(), c = iread(); u++; v++;g[i] = (edge){u, v, w, c};}int l = -101, r = 101, ans = inf;while(l <= r) {int mid = l + r >> 1;int res = check(mid);if(res != -1) ans = res - k * mid, l = mid + 1;else r = mid - 1;}printf("%d\n", ans);return 0;}

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