HDU 4665 Unshuffle DFS找一个可行解

每层找一对相等的整数,分别放在两个不同的串中。

参考了学弟的解法,果断觉得自己老了……

#include <cstdio>
#include <cstring>
#include <cstdlib>

const int MAXN = 2010;

int n;
int num[MAXN];
int vis[MAXN];

bool DFS( int st, int last, int cur )
{
    if ( cur == n / 2 ) return true;

    for ( int i = st; i < n; ++i )
        if ( vis[i] == -1 )
        {
            vis[i] = 0;
            for ( int j = i + 1; j < n; ++j )
                if ( num[j] == num[i] && vis[j] == -1 && j > last )
                {
                    vis[j] = 1;
                    if ( DFS( i + 1, j, cur + 1 ) )
                        return true;
                    vis[j] = -1;
                }
            vis[i] = -1;
            break;
        }
    return false;
}

int main()
{
    int T;
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d", &n );
        for ( int i = 0; i < n; ++i )
            scanf( "%d", &num[i] );

        memset( vis, -1, sizeof(vis) );
        DFS( 0, 0, 0 );
        for ( int i = 0; i < n; ++i )
            printf( "%d", vis[i] );
        puts("");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/GBRgbr/p/3246483.html