HDU - 3496 Watch The Movie(01背包)

题目大意：有N个碟片，你只能选择M个进行播放，最多能观看L分钟。给出每个碟片的播放时间和看完后能收获的快乐值，问能否看完这M个碟片，如果能全部看完的话，输出最大快乐值

解题思路：用dp[i][j]表示i分钟看完j个碟片的最大快乐值，然后背包即可

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110;
const int L = 1010;

int n, m, l;
int Time
, rate
;
int dp[L]
;

void init() {
    scanf("%d%d%d", &n, &m, &l);

    for (int i = 0; i < n; i++) 
        scanf("%d%d", &Time[i], &rate[i]);
}

void solve() {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i <= l; i++) dp[i][0] = 0;

    for (int i = 0; i < n; i++)
        for (int j = l; j >= Time[i]; j--) 
            for (int k = m - 1; k >= 0; k--) 
                if (dp[j - Time[i]][k] != -1)
                    dp[j][k + 1] = max(dp[j][k + 1], dp[j - Time[i]][k] + rate[i]);

    int ans = 0;
    for (int i = 0; i <= l; i++)
        ans = max(dp[i][m], ans);
    printf("%d\n", ans);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}