主題鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4864

Task

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1346????Accepted Submission(s): 336

Problem Description Today the company has m tasks to complete. The ith task need xi minutes to complete. Meanwhile, this task has a difficulty level yi. The machine whose level below this task’s level yi cannot complete this task. If the company completes this task, they will get (500*xi+2*yi) dollars.
The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task can only be completed by one machine.
The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum. Input The input contains several test cases.
The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000).
The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine.
The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task. Output For each test case, output two integers, the maximum number of the tasks which the company can complete today and the money they will get. Sample Input 1 2 100 3 100 2 100 1 Sample Output 1 50004
Author FZU Source 2014 Multi-University Training Contest 1 Recommend We have carefully selected several similar problems for you:??4871?4870?4869?4868?4867?

基本思想是貪心：

對于價值c=500*xi+2*yi，yi最大影響100*2<500,所以就是求xi總和最大。

能夠先對機器和任務的時間從大到小排序。從最大時間的任務開始。找出滿足任務時間要求的全部機器。從中找出等級最低且滿足任務等級要求的機器匹配。依次對任務尋找滿足要求的機器。

代碼例如以下：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int N = 100017; struct work {int x,y; }ma[N],ta[N]; bool cmp(work a, work b) {if(a.x == b.x)return a.y > b.y;return a.x > b.x; } int main() {int n, m;int i, j;int c[N];while(~scanf("%d%d",&n,&m)){memset(c,0,sizeof(c));for(i = 1; i <= n; i++){scanf("%d%d",&ma[i].x,&ma[i].y);}for(i = 1; i <= m; i++){scanf("%d%d",&ta[i].x,&ta[i].y);}sort(ma+1,ma+n+1,cmp);sort(ta+1,ta+m+1,cmp);int l = 1;__int64 ans = 0;int num = 0;for(i = 1; i <= m; i++){while(l <= n&&ma[l].x >= ta[i].x){//尋找全部的能完畢當前任務的機器c[ma[l].y]++;l++;}for(j = ta[i].y; j <= 100; j++){//尋找全部能完畢當前任務的機器中等級最低的if(c[j]){num++;ans+=500*ta[i].x+2*ta[i].y;c[j]--;break;}}}printf("%d %I64d\n",num,ans);}return 0; }

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