暑假集训(2)第八弹 ----- Hero（hdu4310)

时间：2016-07-22 21:18:07 阅读：153 评论：0 收藏：0 [点我收藏+]

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K - Hero

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

1 10 2 2 100 1 1 100

Sample Output

20 201

问题分析：dota啊，好久没玩过了，真......嗯，先统计出所有敌方造成的伤害，然后用dps/hp（每点生命值造成伤害）作为排序的依据，从大到小依次杀光，每杀一个便让总dp减去被杀的人的dp即可。

 1 #include "iostream"
 2 #include "algorithm"
 3 
 4 using namespace std;
 5 struct enemy
 6 {
 7     int hp;
 8     int dp;
 9 };
10 enemy N[21];
11 enemy T[21];
12 int sum;
13 __int64 dead;
14 void ebegin(int n)
15 {
16   for (int i=1;i<=n;i++)
17     {
18       cin>>N[i].dp>>N[i].hp;
19       sum += N[i].dp;
20     }
21 }
22 void defeat (int n)
23 {
24   for (int i=1;i<=n;i++)
25   {
26      dead += sum * N[i].hp;
27       sum -= N[i].dp;
28   }
29    cout<<dead<<endl;
30 }
31 int compare (enemy x,enemy y)
32 {
33     return  x.hp * y.dp < x.dp * y.hp;
34 }
35 int main()
36 {
37     int n;
38     while (cin>>n)
39     {
40         dead = 0;
41         ebegin(n);
42         sort(N+1,N+n+1,compare);
43         defeat(n);
44     }
45     return 0;
46 }

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暑假集训(2)第八弹 ----- Hero（hdu4310)

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原文地址：http://www.cnblogs.com/huas-zlw/p/5697089.html

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