HDU

传送门

限制容量恰好为W的完全背包。

那就设dp[j]为容量为j恰好能得到的最小价值，不存在则设为INF

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;

const int maxn = 5e2 + 5;
const int maxw = 1e4 + 5;
int p_w, tot_w, W;
int N;
int val[maxn], wei[maxn];
int dp[maxw];
int T;

void solve() {
    memset(dp, INF, sizeof(dp));
    dp[0] = 0; //
    for (int i = 0; i < N; i++) {
        for (int j = wei[i]; j <= W; j++) {
            dp[j] = min(dp[j], dp[j - wei[i]] + val[i]);
        }
    }
    if (dp[W] == INF) {
        puts("This is impossible.");
    } else {
        printf("The minimum amount of money in the piggy-bank is %d.
", dp[W]);
    }
}

int main(int argc, const char * argv[]) {
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &p_w, &tot_w);
        W = tot_w - p_w;
        scanf("%d", &N);
        for (int i = 0; i < N; i++) {
            scanf("%d%d", &val[i], &wei[i]);
        }
        solve();
    }
    return 0;
}