POJ 3093 Margaritas on the River Walk 背包DP

dalao讲解

#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-- ) { // 枚举最小的没有被选到的物品
if( m - sum[ i - 1 ] >= 0 ) { //如果现在这些小于的都被选了
int minn = max( m - sum[i] + 1, 0 ); // 找到还剩多少的最小值
//现在至少要把m - sum[i] + 1 的空间全部选满,因为如果不选满,则还剩
//m - ( m - sum[i] + 1 ) = sum[i] - 1;这是剩下的空间,而sum[ i - 1 ]又必须要选,因此这个物品是选不了的
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;
}