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最新推荐文章于 2021-10-14 18:50:15 发布

转载最新推荐文章于 2021-10-14 18:50:15 发布 · 204 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/7286150.html

本文探讨了在Dota游戏中面对强大对手和弱小队友时的策略问题，通过建立英雄属性模型，采用数学公式和算法，计算出最小生命值损失的最优攻击策略。文章深入分析了敌方英雄的伤害值(DPS)和生命值(HP)，并提供了一个算法实现，以确定最佳的战斗顺序。

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.

InputThe 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)OutputOutput one line for each test, indicates the minimum HP loss.Sample Input

Sample Output

20
201

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<cstring>
#include<iostream>
#define MAXN 25
#define INF 0x3f3f3f3f
#define eps 1e-11 + 1e-12/2
typedef long long LL;

using namespace std;
/*
推公式
A的伤害<B的伤害 即
BD*AH <AD*BH
*/
struct node
{
    int d, h;
};
node a[MAXN];
bool cmp(node l, node r)
{
    return    r.d*l.h < r.h*l.d;
}
int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 0; i < n; i++)
            scanf("%d%d", &a[i].d, &a[i].h);
        sort(a, a + n, cmp);
        int time = 0, sum = 0;
        for (int i = 0; i < n; i++)
        {
            time += a[i].h;
            sum += (a[i].d)*time;
        }
        printf("%d\n", sum);
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/7286150.html