HDU

传送门

题意：it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.  If the total number of different values is less than K,just ouput 0.

输入：T组数据 体积V 要求的K    N , V, K(N <= 100 , V <= 1000 , K <= 30)

　　各项价值

　　各项体积

定义dp[i][j][k]为利用前i个和体积j能获得的不重复的第K大值。类似于01背包，可以用二维的dp[j][k]去存储信息。当取第i个物品时，枚举取与不取，得到体积j下的2k种情况，选择其中的k种更新答案即可。就像我有两个数列，要求第K大值，只要分别得知两个数列里的前K个值即可

复杂度是O(NVK)

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

const int maxn = 1e2 + 10;
const int maxv = 1e3 + 10;
const int maxk = 35;

int dp[maxv][maxk];
int slt[maxk], no_slt[maxk];
int val[maxn], vol[maxn];

int N, V, K;

int main(int argc, const char * argv[]) {
    int T; scanf("%d", &T);
    while (T--) {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d%d", &N, &V, &K);
        for (int i = 0; i < N; i++) scanf("%d", &val[i]);
        for (int i = 0; i < N; i++) scanf("%d", &vol[i]);
        for (int i = 0; i < N; i++) {
            for (int j = V; j >= vol[i]; j--) {
                for (int k = 1; k <= K; k++) {
                    slt[k] = dp[j - vol[i]][k] + val[i];
                    no_slt[k] = dp[j][k];
                }
                slt[K + 1] = -1; no_slt[K + 1] = -1;
                int p1 = 1, p2 = 1, p3 = 1;
                while (p1 <= K && (slt[p2] != -1 || no_slt[p3] != -1)) {
                    if (slt[p2] > no_slt[p3]) {
                        dp[j][p1] = slt[p2++];
                    } else {
                        dp[j][p1] = no_slt[p3++];
                    }
                    if (dp[j][p1] != dp[j][p1 - 1])
                        p1++;
                }
                
            }
        }
        printf("%d
", dp[V][K]);
    }
    return 0;
}