本文主要是介绍ZOJ 3209 Treasure Map（DLX精确覆盖），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
ZOJ 3209 Treasure Map
题目链接
题意：给一个大矩形和一些小矩形，问最少几个矩形能覆盖大矩形，不能重复
思路：dlx精确覆盖，以每个矩形个格点为列，以每个小矩形为行，做精确覆盖即可
代码：
#include <cstdio>
#include <cstring>using namespace std;const int MAXNODE = 450005;
const int MAXN = 505;
const int MAXM = 905;const int INF = 0x3f3f3f3f;struct DLX {int n, m, size;int U[MAXNODE], D[MAXNODE], R[MAXNODE], L[MAXNODE], row[MAXNODE], col[MAXNODE];int H[MAXN], S[MAXM];int ansd, ans[MAXN];void init(int n, int m) {this->n = n;this->m = m;ansd = INF;for(int i = 0; i <= m; i++) {S[i] = 0;U[i] = D[i] = i;L[i] = i - 1;R[i] = i + 1;}R[m] = 0; L[0] = m;size = m;for(int i = 1; i <= n; i++)H[i] = -1;}void Link(int r, int c) {++S[col[++size] = c];row[size] = r;D[size] = D[c];U[D[c]] = size;U[size] = c;D[c] = size;if(H[r] < 0) H[r] = L[size] = R[size] = size;else {R[size] = R[H[r]];L[R[H[r]]] = size;L[size] = H[r];R[H[r]] = size;}}void remove(int c) {L[R[c]] = L[c]; R[L[c]] = R[c];for(int i = D[c]; i != c; i = D[i]) {for(int j = R[i]; j != i; j = R[j]) {U[D[j]] = U[j];D[U[j]] = D[j];--S[col[j]];}}}void resume(int c) {for(int i = U[c]; i != c; i = U[i])for(int j = L[i]; j != i; j = L[j])++S[col[U[D[j]] = D[U[j]] = j]];L[R[c]] = R[L[c]] = c;}void Dance(int d) {if(ansd <= d) return;if(R[0] == 0) {if(d < ansd) ansd = d;return;}int c = R[0];for(int i = R[0]; i != 0; i = R[i]) {if(S[i] < S[c])c = i;}remove(c);for(int i = D[c]; i != c; i = D[i]) {for(int j = R[i]; j != i; j = R[j]) remove(col[j]);Dance(d + 1);for(int j = L[i]; j != i; j = L[j]) resume(col[j]);}resume(c);}
} gao;int T, n, m, p;int main() {scanf("%d", &T);while (T--) {scanf("%d%d%d", &n, &m, &p);gao.init(p, n * m);int x1, y1, x2, y2;for (int i = 1; i <= p; i++) {scanf("%d%d%d%d", &x1, &y1, &x2, &y2);for (int x = x1; x < x2; x++) {for (int y = y1; y < y2; y++) {gao.Link(i, x * m + y + 1);}}}gao.Dance(0);if (gao.ansd == INF) gao.ansd = -1;printf("%d\n", gao.ansd);}return 0;
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