poj3740 Easy Finding(深搜)

 Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11432		Accepted: 2802

DescriptionGiven a M×N matrix A. Aij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1. InputThere 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. OutputFor each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line. Sample Input3 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 0Sample OutputYes, I found it
It is impossible
 
 题意 :   给出一个0,1矩阵，一个新矩阵由该矩阵中的若干行组成，问是否存在这样的新矩阵满足每列有且仅有一个1
 
这题深搜加一点简单剪枝水过了，跳舞链有待学习#include <iostream>
using namespace std;

// 对象
int M, N; // 行,列
int m[16][300]; // 原矩阵
int v[300]; // 压缩向量
bool zIndex[16]; // 标记全0行

// 函数
bool Clash( const int row )
{
for( int i = 0; i < N; ++ i )
if( v[i] == 1 && m[row][i] == 1 )
return true; // 冲突
return false;
}
void GetAllZeroRowIndex()
{
memset( zIndex, false, sizeof(zIndex) );
for( int i = 0; i < M; ++ i ) {
bool flag = true;
for( int j = 0; j < N; ++ j ) {
if( m[i][j] == 1 ) {
flag = false;
break;
}
}
if( flag ) zIndex[i] = true; // 标记全0行
}
}
bool Full()
{
for( int i = 0; i < N; ++ i )
if( v[i] == 0 )
return false; // 压缩向量未满
return true;
}
/*
* 深搜, 将第row行加入子矩阵
*/
bool DFS( const int row )
{
// 该行可选
if( !zIndex[row] && !::Clash(row) ) {
for( int i = 0; i < N; ++ i )
v[i] = v[i] | m[row][i];
if( ::Full() )
return true;
}
else // 该行不可选
return false;

// 深搜子问题
for( int i = row + 1; i < M; ++ i )
if( ::DFS(i) )
return true;

// 该行不可行
for( int i = 0; i < N; ++ i )
v[i] = v[i] ^ m[row][i];
return false;
}
// 优化
bool HasColumnAllZero()
{
for( int i = 0; i < N; ++ i ) {
bool flag = true;
for( int j = 0; j < M; ++ j ) {
if( m[j][i] != 0 ) {
flag = false;
break;
}
}
if( flag ) return true; // 存在全0列
}
return false;
}
bool HasRowAllOne()
{
for( int i = 0; i < M; ++ i ) {
bool flag = true;
for( int j = 0; j < N; ++ j ) {
if( m[i][j] != 1 ) {
flag = false;
break;
}
}
if( flag ) return true; // 存在全1行
}
return false;
}

int main()
{
while( scanf( "%d%d", & M, & N ) == 2 ) {
for( int i = 0; i < M; ++ i )
for( int j = 0; j < N; ++ j )
scanf( "%d", & m[i][j] );

bool succ = false;
if( ::HasColumnAllZero() ) succ = false;
else if( ::HasRowAllOne() ) succ = true;
else {
// 标记全0行
::GetAllZeroRowIndex();
memset( v, 0, sizeof(v) );
for( int i = 0; i < M; ++ i )
if( ::DFS(i) ) {
succ = true;
break;
}
}
if( succ ) printf( "Yes, I found it\n" );
else printf( "It is impossible\n" );
}
system( "pause" );
return 0;
}