MUTC 2 A - Hero 状态压缩dp

最新推荐文章于 2021-10-14 18:50:15 发布

cyendra

最新推荐文章于 2021-10-14 18:50:15 发布

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分类专栏： DP 解题报告 Multi-University Training Contest 文章标签：题解 dp

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并结合人工智能与音视频处理的最新进展，展示如何将这些技术应用于游戏开发中，以提升用户体验与创新游戏玩法。

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Hero

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1869 Accepted Submission(s): 868

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

Sample Output

      
       20
201

Author

TJU

Source

2012 Multi-University Training Contest 2

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zhuyuanchen520

---------------

状态压缩dp

f[S]=min(f[S+k]+sum[S+k]*hp[k])

用贪心做的都是异端啊异端！！

---------------

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int maxn=1<<22;
int n;
int dps[22],hp[22];
int sum[maxn];
int f[maxn];

int main()
{
    //f[S]=min(f[S+k]+sum[S+k]*hp[k]);
    while (~scanf("%d",&n))
    {
        memset(f,-1,sizeof(f));
        memset(sum,0,sizeof(sum));

        //init()
        for (int i=0;i<n;i++)
        {
            scanf("%d%d",&dps[i],&hp[i]);
        }
        //prepare()
        for (int i=0;i<(1<<n);i++)
        {
            for (int j=0;j<n;j++)
            {
                if (i&(1<<j))
                {
                    sum[i]+=dps[j];
                }
            }
        }
        //dp
        f[(1<<n)-1]=0;
        for (int s=(1<<n)-1;s>=0;s--)
        {
            for (int k=0;k<n;k++)
            {
                if (!(s&(1<<k)))
                {
                    if (f[s]==-1||f[s+(1<<k)]+sum[s+(1<<k)]*hp[k]<f[s])
                    {
                        f[s]=f[s+(1<<k)]+sum[s+(1<<k)]*hp[k];
                    }
                }
            }
        }
        printf("%d\n",f[0]);
    }
    return 0;
}