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Easy Finding
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16178 | Accepted: 4343 |
Description
Given a M× N matrix A. A ij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.
Input
There are multiple cases ended by EOF. Test case up to 500.The first line of input is M, N ( M ≤ 16, N ≤ 300). The next M lines every line contains N integers separated by space.
Output
For each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.
Sample Input
3 3 0 1 0 0 0 1 1 0 0 4 4 0 0 0 1 1 0 0 0 1 1 0 1 0 1 0 0
Sample Output
Yes, I found it It is impossible
Source
POJ Monthly Contest - 2009.08.23, MasterLuo
解题思路:
题意为由01组成的矩阵,问能不能挑出几行使组成的新矩阵每列只有一个1.
套用Dlx模板,不过G++ 超时,C++勉强能过。
代码:
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxnode=5000;
const int maxm=310;
const int maxn=18;struct DLX
{int n,m,size;int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];int H[maxn];//行头节点int S[maxm];//每列有多少个节点int ansd,ans[maxn];//如果有答案,则选了ansd行,具体是哪几行放在ans[ ]数组里面,ans[0~ansd-1];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;//初始状态下,上下自己指向自己L[i]=i-1;R[i]=i+1;}R[m]=0,L[0]=m;size=m;//编号,每列都有一个头节点,编号1-mfor(int i=1;i<=n;i++)H[i]=-1;//每一行的头节点}void link(int r,int c)//第r行,第c列{++S[Col[++size]=c];//第size个节点所在的列为c,当前列的节点数++Row[size]=r;//第size个节点行位置为rD[size]=D[c];//下面这四句头插法(图是倒着的?)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)//删除节点c,以及c上下节点所在的行,每次调用这个函数,都是从列头节点开始向下删除,这里c也可以理解为第c列{ //因为第c列的列头节点编号为cL[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)//恢复节点c,以及c上下节点所在的行(同上,也可以理解为从第c列的头节点开始恢复{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]]; //打这一行太纠结了 T TL[R[c]]=R[L[c]]=c;}bool dance(int d)//递归深度{if(R[0]==0){ansd=d;return true;}int c=R[0];for(int i=R[0];i!=0;i=R[i])if(S[i]<S[c])c=i;remove(c);//找到节点数最少的列,当前元素不是原图上0,1的节点,而是列头节点for(int i=D[c];i!=c;i=D[i]){ans[d]=Row[i];//列头节点下面的一个节点for(int j=R[i];j!=i;j=R[j])remove(Col[j]);if(dance(d+1))//找到,返回return true;for(int j=L[i];j!=i;j=L[j])resume(Col[j]);}resume(c);return false;}
};DLX x;
int n,m;int main()
{while(scanf("%d%d",&n,&m)!=EOF){x.init(n,m);int num;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>num;if(num)x.link(i,j);}}if(!x.dance(0))printf("It is impossible\n");elseprintf("Yes, I found it\n");}return 0;
}
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