hdu 4502 吉哥系列故事——临时工计划

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题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=4502

本题是动态规划，dp[i]表示到今天所能挣到最多的钱，dp[i] = MAX(dp[i], dp[j] + a[j+1][i]);其中a[i][j]表示：从i到j天的工资！

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN  = 110;int MAX(int a, int b){    return a > b ? a : b;}int main(){    int T, iHoliday, iWorks, i, j, iStart, iEnd, iSalary;    int iWorkDay[MAXN][MAXN];    int dp[MAXN];    scanf("%d", &T);    while(T--)    {        scanf("%d %d", &iHoliday, &iWorks);        memset(iWorkDay, 0, sizeof(iWorkDay));        memset(dp, 0, sizeof(dp));        for(i = 0; i < iWorks; ++i)        {            scanf("%d %d %d", &iStart, &iEnd, &iSalary);            if(iStart > iHoliday || iEnd > iHoliday)                continue ;            if(iSalary > iWorkDay[iStart][iEnd])                iWorkDay[iStart][iEnd] = iSalary;        }        for(i = 1; i <= iHoliday; ++i)        {            for(j = 0; j < i; ++j)            {                dp[i] = MAX(dp[i], dp[j] + iWorkDay[j+1][i]);            }        }        printf("%d\n", dp[iHoliday]);    }    return 0;}

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