HDU1789 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;
}