题意:有一个n * n的棋盘,每个棋盘有某些矩形区域被染成了黑色(这些矩形区域有可能相交),问把所有黑色区域染成白色的最小花费是多少?你每次可以选择把一个矩形区域染成白色,花费是染色的矩形区域长和宽的最小值。
思路:容易发现,假设一个矩形的坐标是(l1, r1, l2, r2),假设(l2 - l1 < r2 - r1), 那么我们把r1和r2变成1和n对答案不会有影响。那么,我们发现问题转化为了选最少的行和列使得所有的黑色区域被覆盖,换句话说,就是所有的点至少被一个行货列所覆盖。我们把行看成左边的点,列看成右边的点,黑色的点看成边,那么这个问题就变成了求二分图的最小点覆盖。但是直接用二分图做边数点数会很大,所有我们可以离散化之后变成求网络流的最大流。
代码:
#include <bits/stdc++.h> #define pii pair<int, int> #define LL long long #define INF 1e18 using namespace std; const int maxm = 100100; const int maxn = 510; bool v[310][310]; int head[maxn], Next[maxm * 6], ver[maxm * 6], tot = 1; int d[maxn], s ,t; LL edge[maxm * 6]; void add(int x, int y, LL z) { ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot; ver[++tot] = x, edge[tot] = 0, Next[tot] = head[y], head[y] = tot; } queue<int> q; bool bfs() { memset(d, 0, sizeof(d)); while(q.size()) q.pop(); q.push(s), d[s] = 1; while(q.size()) { int x = q.front(); q.pop(); for (int i = head[x]; i; i = Next[i]) { if(edge[i] && !d[ver[i]]) { q.push(ver[i]); d[ver[i]] = d[x] + 1; if(ver[i] == t) return 1; } } } return 0; } LL dinic(int x, LL flow) { if(x == t) return flow; LL rest = flow, k; for (int i = head[x]; i && rest; i = Next[i]) { if(edge[i] && d[ver[i]] == d[x] + 1) { k = dinic(ver[i], min(rest, edge[i])); if(!k) d[ver[i]] = 0; edge[i] -= k; edge[i ^ 1] += k; rest -= k; } } return flow - rest; } vector<int> X, Y; vector<pii> L, R; int main() { int n, m, l1, r1, l2, r2, sz; scanf("%d%d", &n, &m); X.push_back(0), X.push_back(n); Y.push_back(0), Y.push_back(n); for (int i = 1; i <= m; i++) { scanf("%d%d%d%d", &l1, &r1, &l2, &r2); l1--, r1--; L.push_back(make_pair(l1, r1)); R.push_back(make_pair(l2, r2)); X.push_back(l1), X.push_back(l2); Y.push_back(r1), Y.push_back(r2); } sort(X.begin(), X.end()); sort(Y.begin(), Y.end()); sz = unique(X.begin(), X.end()) - X.begin(); while(X.size() > sz) X.pop_back(); sz = unique(Y.begin(), Y.end()) - Y.begin(); while(Y.size() > sz) Y.pop_back(); s = 210, t = 211; for (int i = 0; i < X.size() - 1; i++) { add(s, i, X[i + 1] - X[i]); } for (int i = 0; i < Y.size() - 1; i++) { add(i + X.size(), t, Y[i + 1] - Y[i]); } for (int i = 0; i < m; i++) { l1 = lower_bound(X.begin(), X.end(), L[i].first) - X.begin(); r1 = lower_bound(Y.begin(), Y.end(), L[i].second) - Y.begin(); l2 = lower_bound(X.begin(), X.end(), R[i].first) - X.begin(); r2 = lower_bound(Y.begin(), Y.end(), R[i].second) - Y.begin(); for (int i = l1; i < l2; i++) { for (int j = r1; j < r2; j++) { v[i][j] = 1; } } } for (int i = 0; i < X.size() - 1; i++) { for (int j = 0; j < Y.size() - 1; j++) { if(v[i][j]) { add(i, j + X.size(), INF); } } } LL flow = 0, maxflow = 0; while(bfs()) while(flow = dinic(s, INF)) maxflow += flow; printf("%lld ", maxflow); }