HDOJ 4310 —— 排序

Hero

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2251    Accepted Submission(s): 1012

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

   
   
   
   

    
    
    
    1
10 2
2
100 1
1 100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    20
201
   
   
   
   

 

Author

TJU

 

Source

2012 Multi-University Training Contest 2

 

Recommend

zhuyuanchen520

 

就是一个排序。。。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
//#define mid ((l + r) >> 1)
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 50;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x3f3f3f3f;
const int IMIN = 0x80000000;
const double E = 2.718281828;
#define eps 1e-8
#define DEBUG 1
#define mod 100000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
//    #pragma comment(linker, "/STACK:102400000,102400000")__int64 a[10050];
struct Node
{
    double dps , hp;
    bool operator < (const Node & a)const
    {
        return dps / hp > a.dps / a.hp;
    }

}node[MAXN];
int main()
{
    int n ;
    while(~scanf("%d"  ,&n))
    {
        double sum = 0.0;
        FOR(i , 0 , n)
        {
            scanf("%lf%lf" , &node[i].dps , &node[i].hp);
            sum += node[i].dps;
        }
        sort(node , node + n);
        int ans = 0;
        FOR(i ,0 , n)
        {
            ans += (int)node[i].hp * sum;
            sum -= node[i].dps;
        }
        printf("%d\n" , ans);
    }
    return 0;
}