UVA10891Game of Sum

Posted 2020-08-21

题解:

这个题的题意挺误解人的。

每次选取的最优策略。我的理解每次从左开始。或从右开始。选取最大值(每个位置选取的最大值的位置是固定的).把剩下的序列给对方....但题意并不是这样

看了题解。

表示dp[ i ][ j ] 表示 区间 i j 先选得到的最大值.

dp[ i ][ j ] = sum( j ) - sum( i - 1 ) - min( dp( i + 1, j ), dp( i + 2, j ), dp( i + 3, j )...dp( j ,j ), dp(  i , j - 1 ), dp( i ,j - 2 ).....dp( i ,i ), 0 );

o( n ^ 3 ) 和o( n ^ 2 )的优化书上写得很清楚了,不多说

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
const int INF = 1000000000;
const int maxn = 105;


int n;
int a[ maxn ], sum[ maxn ];
int dp[ maxn ][ maxn ], vis[ maxn ][ maxn ];

int DP( int l, int r )
{
    if( vis[ l ][ r ] ) return dp[ l ][ r ];
    vis[ l ][ r ] = 1;
    int m = 0;
    for( int i = l + 1; i <= r; i ++ ) m = min( m, DP( i, r ) );
    for( int i = r - 1; i >= l; i -- ) m = min( m, DP( l ,i ) );
    return dp[ l ][ r ] = sum[ r ] - sum[ l - 1 ] - m;
}

int main()
{
    while( ~scanf( "%d", &n ) && n )
    {
        for( int i = 1; i <= n; i ++ )
        {
            scanf( "%d", &a[ i ] );
            sum[ i ] = sum[ i - 1 ] + a[ i ];
        }
        memset( vis, 0, sizeof( vis ) );
        printf( "%d\n", DP( 1 , n ) * 2 - sum[ n ] );
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
const int INF = 1000000000;
const int maxn = 105;


int n;
int a[ maxn ], sum[ maxn ];
int dp[ maxn ][ maxn ], f[ maxn ][ maxn ], g[ maxn ][ maxn ];

int main()
{
    while( ~scanf( "%d", &n ) && n )
    {
        for( int i = 1; i <= n; i ++ )
        {
            scanf( "%d", &a[ i ] );
            sum[ i ] = sum[ i - 1 ] + a[ i ]; 
        }
        for( int i = 1; i <= n; i ++ )  f[ i ][ i ] = g[ i ][ i ] = dp[ i ][ i ] = a[ i ];
        for( int k = 1; k < n; k ++ )
        for( int i = 1; i + k <= n; i ++ )
        {
            int j = i + k;
            int m = 0;
            m = min( m, f[ i + 1 ][ j ] );
            m = min( m, g[ i ][ j - 1 ] );
            dp[ i ][ j ] = sum[ j ] - sum[ i - 1 ] - m;
            f[ i ][ j ] = min( f[ i + 1 ][ j ], dp[ i ][ j ] );
            g[ i ][ j ] = min( g[ i ][ j - 1 ], dp[ i ][ j ] );
        }
        printf( "%d\n", 2 * dp[ 1 ][ n ] - sum[ n ] );
    }
    return 0;
}