hdu 1978 简单dp

来源：互联网发布：所有股票数据编辑：程序博客网时间：2024/05/16 06:03

其实就是枚举当前点，然后枚举可以到打的点，将初始点初始为１，然后每个可到达的点加上当前点的可到达数

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int dp[107][107];int a[107][107];int main ( ){    int t,n,m;    scanf ( "%d" , &t );    while ( t-- )    {        scanf ( "%d%d" , &n , &m );        for ( int i = 1 ; i <= n ; i++ )            for ( int j = 1 ; j <= m ; j++ ) scanf ( "%d" , &a[i][j] );        memset ( dp , 0 , sizeof ( dp ) );        dp[1][1] = 1;        for ( int i = 1 ; i <= n ; i++ )            for ( int j = 1 ; j <= m ; j++ )                for ( int k = 0 ; k <= a[i][j] ; k++ )                    for ( int t = 0 ; t <= a[i][j] - k ; t++ )                    {                        if ( k == t && k == 0 ) continue;                        if ( i + k > n ) continue;                        if ( j + t > m ) continue;                        dp[i+k][j+t] += dp[i][j];                        dp[i+k][j+t] %= 10000;                    }        printf ( "%d\n" , dp[n][m] );    }}

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