POJ 3093 Margaritas on the River Walk 揹包DP

dalao講解

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因此我們可以揹包，按照枚舉最小的一個不被選中的物品進行巧妙去重

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm> 
using namespace std;
const int MAXN = 10000 + 10;
int T, f[MAXN], n, m, sum[MAXN], w[MAXN];
int main( ) {
    scanf( "%d", &T );
    for( register int kase = 1; kase <= T; kase++ ) {
        memset( f, 0, sizeof(f) );
        memset( sum, 0, sizeof(sum) );
        f[0] = 1;
        scanf( "%d%d", &n, &m );
        for( register int i = 1; i <= n; i++ ) scanf( "%d", &w[i] );
        sort( w + 1, w + n + 1 );
        if( w[1] > m ) { printf( "%d 0\n", kase ); continue; }
        for( register int i = 1; i <= n; i++ ) sum[i] = sum[ i - 1 ] + w[i];
        int ans = 0;
        for( register int i = n; i >= 1; i-- ) { // 枚舉最小的沒有被選到的物品 
            if( m - sum[ i - 1 ] >= 0 ) { //如果現在這些小於的都被選了 
                int minn = max( m - sum[i] + 1, 0 ); // 找到還剩多少的最小值
                //現在至少要把m - sum[i] + 1 的空間全部選滿,因爲如果不選滿,則還剩
                //m - ( m - sum[i] + 1 ) = sum[i] - 1;這是剩下的空間,而sum[ i - 1 ]又必須要選,因此這個物品是選不了的 
                for( register int j = minn; j <= m - sum[ i - 1 ]; j++ ) ans += f[j];
            }
            for( register int j = m; j >= w[i]; j-- ) f[j] += f[ j - w[i] ];
        }
        printf( "%d %d\n", kase, ans );
    }
    return 0;
}