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. Then p lines follow, each consists of four integers x1 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.
Author: HANG, Hang
Source: The 6th Zhejiang Provincial Collegiate Programming Contest
DancingLinksX的題目
題意給你一些碎的地圖
然後你知道這個碎塊屬於大地圖的哪部分
讓你用其中的儘量少的一些拼出大地圖
碎塊作爲行，大地圖的每一個小的單元格作爲列
建圖即可
這裏注意dance即dfs一定要減掉超過目前最優解的情況
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MN=505;
const int MM=35*35;
const int MNN=MN*MM+MM; //最大點數
struct DLX
{
int n,m,si;//n行數m列數si目前有的節點數
//十字鏈表組成部分
int U[MNN],D[MNN],L[MNN],R[MNN],Row[MNN],Col[MNN];
//第i個結點的U向上指針D下L左R右，所在位置Row行Col列
int H[MN],S[MM]; //記錄行的選擇情況和列的覆蓋情況
int ansd,ans[MN];
void init(int _n,int _m) //初始化空表
{
n=_n;
m=_m;
for(int i=0;i<=m;i++) //初始化第一橫行（表頭）
{
S[i]=0;
U[i]=D[i]=i; //目前縱向的鏈是空的
L[i]=i-1;
R[i]=i+1; //橫向的連起來
}
R[m]=0;L[0]=m;
si=m; //目前用了前0~m個結點
for(int i=1;i<=n;i++)
H[i]=-1;
}
void link(int r,int c) //插入點(r,c)
{
//cout<<"link "<<r<<" "<<c<<endl;
++S[Col[++si]=c]; //si++;Col[si]=c;S[c]++;
Row[si]=r;
D[si]=D[c];
U[D[c]]=si;
U[si]=c;
D[c]=si;
if(H[r]<0)
H[r]=L[si]=R[si]=si;
else
{
R[si]=R[H[r]];
L[R[H[r]]]=si;
L[si]=H[r];
R[H[r]]=si;
}
}
void remove(int c) //列表中刪掉c列
{
L[R[c]]=L[c];//表頭操作
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) //恢復c列
{
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) //選取了d行
{
if(ansd!=-1&&d>=ansd) //特別注意這裏剪枝
return;
if(R[0]==0)//全部覆蓋了
{
//全覆蓋了之後的操作
//if(ansd==-1) 能進來的d值顯然符合上下兩個條件
ansd=d;
//else 但是不註釋掉這些速度更快……
// ansd=min(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])
{
//ans[d]=Row[i];
for(int j=R[i];j!= i;j=R[j])
remove(Col[j]);
dance(d+1);
//cout<<"R0 "<<R[0]<<endl;
for(int j=L[i];j!=i;j=L[j])
resume(Col[j]);
}
resume(c);
return;
}
}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);
for(int ii=1;ii<=p;ii++)
{
int x,xx,y,yy;
scanf("%d%d%d%d",&x,&y,&xx,&yy);
for(int i=x+1;i<=xx;i++)
for(int j=y+1;j<=yy;j++)
dlx.link(ii,(i-1)*m+j);
}
dlx.ansd=-1;
dlx.dance(0);
printf("%d\n",dlx.ansd);
}
return 0;
}