【Dance Links X】
Dance Links X 舞蹈链算法， 原理为一个双端十字链表的数据结构， 主要用于处理精确覆盖问题（解决n皇后问题， 数独问题）。
下面的代码为 【ZOJ-3209 Treasure Map】的题解， 此题要求的是整个面积的覆盖， 把每个格子当成一个列， 然后用模板跑一次就行。
#include <bits/stdc++.h>using namespace std;const int maxnode = 5e5 + 10;
const int MAXM = 10010;
const int MAXN = 510;struct DLX
{int n, m, size;int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];int H[MAXM], S[MAXM];int ansd;void init(int _n, int _m){n = _n;m = _m;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) {//jian zhiif (ansd != -1 && ansd <= d) return;if (R[0] == 0) {if (ansd == -1) ansd = d;else 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);}
};
DLX g;
int main()
{int t, n, m, p;cin >> t;while (t--) {cin >> n >> m >> p;g.init(p, n*m);int x1, x2, y1, y2;for (int k = 1; k <= p; k++) {cin >> x1 >> y1 >> x2 >> y2;for (int i = x1 + 1; i <= x2; i++) {for (int j = y1 + 1; j <= y2; j++) {g.Link (k, j + (i - 1)*m);}}}g.ansd = -1;g.Dance (0);cout << g.ansd << "\n";}return 0;
}