HDU 4310 Hero

Hero

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4912    Accepted Submission(s): 2153

Problem Description

When playing DotA with god-like rivals and pig-like team members, you have to face an embarrassing situation: All your teammates are killed, and you have to fight 1vN.

There are two key attributes for the heroes in the game, health point (HP) and damage per shot (DPS). Your hero has almost infinite HP, but only 1 DPS.

To simplify the problem, we assume the game is turn-based, but not real-time. In each round, you can choose one enemy hero to attack, and his HP will decrease by 1. While at the same time, all the lived enemy heroes will attack you, and your HP will decrease by the sum of their DPS. If one hero's HP fall equal to (or below) zero, he will die after this round, and cannot attack you in the following rounds.

Although your hero is undefeated, you want to choose best strategy to kill all the enemy heroes with minimum HP loss.

 

Input

The first line of each test case contains the number of enemy heroes N (1 <= N <= 20). Then N lines followed, each contains two integers DPSi and HPi, which are the DPS and HP for each hero. (1 <= DPSi, HPi <= 1000)

 

Output

Output one line for each test, indicates the minimum HP loss.

 

Sample Input

110 22100 11 100

 

Sample Output

20201

 

Author

TJU

 

Source

2012 Multi-University Training Contest 2

 

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贪心题，题意：

你是一个不死的打手，每次只能攻击一人，能打掉对方一滴血。在你被围殴时，你要用最少的攻击打败这些人。dps是对方的攻击力，hp是对方的血量。为了让你的攻击次数最少，需要从对方dps/hp最大的开始打，因为他打人最疼而且血相对厚，当对方dps/hp相同时，那就从血最少的开始打，容易打死。

如： 100 1   

         1 100

这组数据，显然对dps/hp排序仍然是这个顺序。于是先打dps100的，要打一次打死，在此期间你会受到来自第一个人100点和第二个人1点攻击，加起来101。下一次攻击第一个人狗带了，只有第二个人打你，一次1点攻击但是100点血，所以你要打他100次打死。101+100=201，所以此组数据结果是201。

这题测试的时候我没看懂，白玩了这么多年游戏_(:з」∠)_

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;struct game{double dps,hp;}num[100];double cmp(const game &a,const game &b){if(a.dps/a.hp!=b.dps/b.hp)return a.dps/a.hp>b.dps/b.hp;else return a.hp<b.hp;}int main(){int n,i;double sum,ans;while(~scanf("%d",&n)){memset(num,0,sizeof(num));sum=0;for(i=0;i<n;i++){scanf("%lf%lf",&num[i].dps,&num[i].hp);sum+=num[i].dps;}sort(num,num+n,cmp);ans=0;for(i=0;i<n;i++){ans+=sum*num[i].hp;sum-=num[i].dps;}printf("%.0lf\n",ans);}return 0;}