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2015
05-23

HDU 4310-Hero-贪心-[解题报告]HOJ

Hero

问题描述 :

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.

输入:

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)

输出:

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)

样例输入:

1
10 2
2
100 1
1 100

样例输出:

20
201

/*
分析:
    简单贪心。
    我来个擦,难道当真是流年不利、今年不适合我做题?怎么这两天
老是看到一个题,很快想到正确思路,然后老是不小心手一抖,就在关
键地方犯致命的错误,什么减号写成小于号啦。。。
    思路很简单,和今年金华的A题是一样的,对E[i].t*E[l].v进行排
序就行了。

                                                              2012-12-12
*/

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
struct A
{
	int t,v;
}E[30];
int cmp(const void *a,const void *b)
{
	A *c,*d;
	c=(A *)a;
	d=(A *)b;
	return (c->t*d->v)-(d->t*c->v);
}

int main()
{
	int n;
	int i,l;
	int base,ans;
	while(scanf("%d",&n)!=-1)
	{
		base=0;
		for(i=0;i<n;i++)	{scanf("%d%d",&E[i].v,&E[i].t);base+=E[i].v;}
		qsort(E,n,sizeof(E[0]),cmp);

		ans=0;
		for(i=0;i<n;i++)
		{
			ans+=base*E[i].t;
			base-=E[i].v;
		}
		printf("%d\n",ans);
	}
	return 0;
}

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参考:http://blog.csdn.net/ice_crazy/article/details/8286908