ZOJ 3209 Treasure Map DLX精确覆盖

来源：互联网发布：2016年nba总决赛g4数据编辑：程序博客网时间：2024/05/16 17:15

题目：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3209

题意：

有一个n*m的矩形，现在有p个碎片，每个碎片精确的覆盖某个区域，问最少需要多少碎片可以恰好覆盖这个矩形，要求任意碎片之间不能有重叠的部分

思路：

可以发现矩形的范围比较小，那么把矩形内的点都抽象出来，碎片内的点都抽象出来，碎片内的覆盖对应的矩形中的点，就是精确覆盖问题了。

#include <bits/stdc++.h>using namespace std;const int X = 500000 + 10, N = 500 + 10, M = 1000 + 10, INF = 0x3f3f3f3f;int res;struct DLX{    int U[X], D[X], L[X], R[X], row[X], col[X];    int H[N], S[M];    int head, sz, tot, n, m, ans[N];    void init(int _n, int _m)    {        n = _n, m = _m;        for(int i = 0; i <= m; i++)            L[i] = i-1, R[i] = i+1, U[i] = D[i] = i, S[i] = 0;        head = 0, tot = 0, sz = m;        L[head] = m, R[m] = head;        for(int i = 1; i <= n; i++) H[i] = -1;    }    void link(int r, int c)    {        ++S[col[++sz]=c];        row[sz] = r;        D[sz] = D[c], U[D[c]] = sz;        U[sz] = c, D[c] = sz;        if(H[r] < 0) H[r] = L[sz] = R[sz] = sz;        else R[sz] = R[H[r]], L[R[H[r]]] = sz, L[sz] = H[r], R[H[r]] = sz;    }    void del(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])                D[U[j]] = D[j], U[D[j]] = U[j], --S[col[j]];    }    void recover(int c)    {        for(int i = U[c]; i != c; i = U[i])            for(int j = L[i]; j != i; j = L[j])                D[U[j]] = U[D[j]] = j, ++S[col[j]];        R[L[c]] = L[R[c]] = c;    }    void dance(int dep)    {        if(R[head] == head)        {            res = min(res, dep-1); return;        }        if(dep-1 >= res) return;        int c = R[head];        for(int i = R[head]; i != head; i = R[i])            if(S[i] < S[c]) c = i;        del(c);        for(int i = D[c]; i != c; i = D[i])        {            ans[dep] = row[i];            for(int j = R[i]; j != i; j = R[j]) del(col[j]);            dance(dep + 1);            for(int j = L[i]; j != i; j = L[j]) recover(col[j]);        }        recover(c);    }}dlx;int main(){    int t, n, m, p;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &n, &m, &p);        dlx.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 j = x1+1; j <= x2; j++)                for(int k = y1+1; k <= y2; k++)                {                    dlx.link(i, (j-1)*m + k);                }        }        res = INF;        dlx.dance(1);        if(res == INF) res = -1;        printf("%d\n", res);    }    return 0;}

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