[UVA][dp] 672 - Gangsters＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
49的鼓 6站台

2014-01-24 09:40:30| 人1,398| 回0 | 上一篇 | 下一篇

[UVA][dp] 672 - Gangsters

推 0 收藏 0 0 站台

Gangsters

N gangsters are going to a restaurant. The i-th gangster comes at thetime T_i and has the prosperity P_i. The door of the restaurant hasK+1 states of openness expressed by the integers in the range [0, K]. Thestate of openness can change by one in one unit of time; i.e. it either opensby one, closes by one or remains the same. At the initial moment of time thedoor is closed (state 0). The i-th gangster enters the restaurant only if thedoor is opened specially for him, i.e. when the state of openness coincideswith his stoutness S_i. If at the moment of time when the gangstercomes to the restaurant the state of openness is not equal to his stoutness,then the gangster goes away and never returns.

The restaurant works in the interval of time [0, T].

The goal is to gather the gangsters with the maximal total prosperity in the restaurant by opening and closing the door appropriately.

Input

The first line of the input is an integer M, then a blank line followed by M datasets. There is a blank line between datasets.

The first line of each dataset contains the values N, K, and T,separated by spaces. ()

The second line of the dataset contains the moments of time whengangsters come to the restaurant ,separated byspaces. (for )

The third line of the dataset contains the values of the prosperity ofgangsters ,separated by spaces. (for )

The forth line of the dataset contains the values of the stoutness ofgangsters ,separated by spaces. (for)

All values in the input file are integers.

Output

For each dataset, print the single integer - the maximal sum of prosperity of gangsters in the restaurant. In case when nogangster can enter the restaurant the output should be 0.Print a blank line between datasets.

Sample Input

14 10 2010 16 8 1610 11 15 110 7 1 8/font>

Sample Output

Miguel Revilla
2000-05-22

目描述：

一家餐在有 N 位流氓，家店的很特殊，每刻可以大+1不+0小-1，

第 0 刻大小 0，而每流氓，如果大小好等於他所要求的，他便消。

反之否，求最大消。

目解法：

作 DP，dp[i][j] : 在 i 大小 j 的最大消。

// 一始想排序，把 i 替成第 i 人，由於可能重，理。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct Gangster {
    int t, p, s;
};
int dp[30005][105], w[30005][105];
int main() {
    int testcase;
    int N, K, T;
    int i, j, k;
    Gangster D[105];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d", &N, &K, &T);
        for(i = 1; i <= N; i++)    scanf("%d", &D[i].t);
        for(i = 1; i <= N; i++)    scanf("%d", &D[i].p);
        for(i = 1; i <= N; i++)    scanf("%d", &D[i].s);
        memset(w, 0, sizeof(w));
        memset(dp, 0, sizeof(dp));
        for(i = 1; i <= N; i++)
            w[D[i].t][D[i].s] += D[i].p;
        int ret = 0;
        for(i = 0; i < T; i++) {
            for(j = min(i, K); j >= 0; j--) {
                if(j)
                    dp[i+1][j-1] = max(dp[i+1][j-1], dp[i][j] + w[i+1][j-1]);
                dp[i+1][j] = max(dp[i+1][j], dp[i][j] + w[i+1][j]);
                dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + w[i+1][j+1]);
            }
        }
        for(i = 0; i <= K; i++)
            ret = max(ret, dp[T][i]);
        printf("%d\n", ret);
        if(testcase)
            puts("");
    }
    return 0;
}