HDU-1789 Doing Homework again 动态规划 Or 贪心

http://acm.hdu.edu.cn/showproblem.php?pid=1789

题义是给定一个作业序列，求如何分配使得得到的分数最多。设 dp[i][j] 代表截止到第i个作业，第j天所能够完成的最多分数。

dp方程为：

if (1<= j <= e[i].time)     dp[i][j] = max(dp[i-1][j], d[i-1][j-1]+e[i].score);

if (e[i].time+1 <= j <= Last)    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);

代码如下：

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <algorithm>

#define MAXN 1005

using namespace std;



int N, sum, Last, dp[MAXN][MAXN];



struct Node

{

    int time;

    int score; 

}e[MAXN];



bool cmp(Node a, Node b)

{

    if (a.time != b.time) {

        return a.time < b.time;

    }

    else {

        return b.score < a.score;

    }

}



int DP()

{

    sort(e+1, e+1+N, cmp);

    for (int i = 1; i <= N; ++i) {

        for (int j = 1; j <= e[i].time; ++j) {

            dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]+e[i].score);

        }

        for (int j = e[i].time+1; j <= Last; ++j) {

            dp[i][j] = max(dp[i-1][j], dp[i][j-1]);

        }

    }

    return dp[N][Last];

}



void init()

{

    for (int i = 1; i <= N; ++i) {

        for (int j = 1; j <= Last; ++j) {

            dp[i][j] = 0;

        }

    }

}



int main()

{

    int T;

    scanf("%d", &T);

    while (T--) {

        sum = Last = 0;

        scanf("%d", &N); 

        for (int i = 1; i <= N; ++i) {

            scanf("%d", &e[i].time);

            Last = max(Last, e[i].time);

        }

        for (int i = 1; i <= N; ++i) {

            scanf("%d", &e[i].score);

            sum += e[i].score;

        }

        init();

        printf("%d\n", sum - DP());

    }

    return 0;

}

 

还有一种思路是利用贪心来做，为什么可以应用贪心规则呢？因为在这个题中，得分高的任务和得分低的都只占用1天的时间，这一天给谁不是给啊，还不如给得分高的，并把这个任务安排在限定期的最后一天。

代码如下：

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#define MAXN 1005

using namespace std;



int N, sum, hash[MAXN];



struct Node

{

    int time, score;

}e[MAXN];



bool cmp(Node a, Node b)

{

    if (a.score != b.score) {

        return a.score > b.score;

    }

    else {

        return a.time < b.time;  

    }

}



int Greedy()

{

    int ct, total = 0;

    sort(e+1, e+1+N, cmp);

    for (int i = 1; i <= N; ++i) {

        ct = e[i].time;

        while (hash[ct]) {

            --ct;  // 无论如何都让这件工作做完

        }

        if (ct > 0) {

            total += e[i].score;

            hash[ct] = 1;

        }

    }

    return total;

}



int main()

{

    int T;

    scanf("%d", &T);

    while (T--) {

        sum = 0;

        memset(hash, 0, sizeof (hash));

        scanf("%d", &N);

        for (int i = 1; i <= N; ++i) {

            scanf("%d", &e[i].time);

        }    

        for (int i = 1; i <= N; ++i) {

            scanf("%d", &e[i].score);

            sum += e[i].score;

        }

        printf("%d\n", sum - Greedy());

    }

    return 0;

}