ZOJ3209Treasure Map精确覆盖DLX

 Treasure Map

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Time Limit: 2 Seconds Memory Limit: 32768 KB

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Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luckily, it is possible to figure out the position of each piece in the original map. Now the boss asks you, the talent programmer, to make a complete treasure map with these pieces. You need to make only one complete map and it is not necessary to use all the pieces. But remember, pieces are not allowed to overlap with each other (See sample 2).

Input

The first line of the input contains an integer T (T <= 500), indicating the number of cases.

For each case, the first line contains three integers n m p (1 <= n, m <= 30, 1 <= p <= 500), the width and the height of the map, and the number of pieces. Thenp lines follow, each consists of four integersx1 y1 x2 y2 (0 <=x1 <x2 <=n, 0 <=y1 <y2 <= m), where (x1, y1) is the coordinate of the lower-left corner of the rectangular piece, and (x2, y2) is the coordinate of the upper-right corner in the original map.

Cases are separated by one blank line.

Output

If you can make a complete map with these pieces, output the least number of pieces you need to achieve this. If it is impossible to make one complete map, just output -1.

Sample Input

3

5 5 1

0 0 5 5

5 5 2

0 0 3 5

2 0 5 5

30 30 5

0 0 30 10

0 10 30 20

0 20 30 30

0 0 15 30

15 0 30 30

Sample Output

1

-1

2

Hint

For sample 1, the only piece is a complete map.

For sample 2, the two pieces may overlap with each other, so you can not make a complete treasure map.

For sample 3, you can make a map by either use the first 3 pieces or the last 2 pieces, and the latter approach one needs less pieces.

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Author: HANG, Hang
Source: The 6th Zhejiang Provincial Collegiate Programming Contest
SubmitStatus 

题目意思是给你一个n*m的矩形，有p个小矩形，问你最少需要用多少个小矩形不重复的拼出n*m的矩形。

很裸的精确覆盖用DLX解决，但是我WA了无数次。。。

建图：p个小矩形每个一行，每行n*m列，每一列i*m+j的格子被这个小矩形覆盖

我的错误：

1.列是i*m+j，我写成i*n+j

2.dfs回溯求最小值，我写成求一次就返回了。。。

 网上的代码基本是用静态数组的，我用的是静态链表，但是效率比他们高。

代码:

Run IDSubmit TimeJudge StatusProblem IDLanguageRun Time(ms)Run Memory(KB)User Name26442802011-08-30 14:34:16Accepted 3209C++1801156魔神翼

#include<cstdio>#define N 505#define M 1005int m,n,H,cnt,size[M],ans;struct Node{    int r,c;    Node *U,*D,*L,*R;}node[40005],row[N],col[M],head;void init(int r,int c){    cnt=0;    head.r=r;    head.c=c;    head.L=head.R=head.U=head.D=&head;    for(int i=0;i<c;i++){        col[i].r=r;        col[i].c=i;        col[i].L=&head;        col[i].R=head.R;        col[i].U=col[i].D=col[i].L->R=col[i].R->L=&col[i];        size[i]=0;    }    for(int i=r-1;i>=0;i--){        row[i].r=i;        row[i].c=c;        row[i].U=&head;        row[i].D=head.D;        row[i].L=row[i].R=row[i].U->D=row[i].D->U=&row[i];    }}void insert(int r,int c){    Node *p=&node[cnt++];    p->r=r;    p->c=c;    p->R=&row[r];    p->L=row[r].L;    p->L->R=p->R->L=p;    p->U=&col[c];    p->D=col[c].D;    p->U->D=p->D->U=p;    ++size[c];}void delLR(Node *p){    p->L->R=p->R;    p->R->L=p->L;}void delUD(Node *p){    p->U->D=p->D;    p->D->U=p->U;}void resumeLR(Node *p){p->L->R=p->R->L=p;}void resumeUD(Node *p){p->U->D=p->D->U=p;}void cover(int c){    if(c==H)        return;Node *R,*C;    delLR(&col[c]);    for(C=col[c].D;C!=&col[c];C=C->D)for(R=C->L;R!=C;R=R->L){--size[R->c];delUD(R);}}void resume(int c){    if(c==H)        return;    Node *R,*C;    for(C=col[c].U;C!=&col[c];C=C->U)for(R=C->R;R!=C;R=R->R){++size[R->c];resumeUD(R);}    resumeLR(&col[c]);}void dfs(int k){    if(head.L==&head){if(k<ans)ans=k;return;}if(k>=ans)return;    int INF=-1u>>1,c=-1;Node *p,*rc;    for(p=head.L;p!=&head;p=p->L)        if(size[p->c]<INF)            INF=size[c=p->c];if(!INF)return;    cover(c);    for(p=col[c].D;p!=&col[c];p=p->D){        for(rc=p->L;rc!=p;rc=rc->L)            cover(rc->c);        dfs(k+1);        for(rc=p->R;rc!=p;rc=rc->R)            resume(rc->c);    }    resume(c);}int main(){int t,p;scanf("%d",&t);    while(t--){int i,j,k,x0,x1,y0,y1;scanf("%d%d%d",&n,&m,&p);init(p+1,H=n*m);for(k=0;k<p;k++){scanf("%d%d%d%d",&x0,&y0,&x1,&y1);for(i=x0;i<x1;i++)for(j=y0;j<y1;j++)insert(k,i*m+j);}ans=999;dfs(0);if(ans<999)printf("%d\n",ans);elseputs("-1");}}