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  • hdu 2616 Kill the monster(stl全排列+暴力)

    Kill the monster

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 502    Accepted Submission(s): 345


    Problem Description
    There is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it.
    Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.
     
    Input
    The input contains multiple test cases.
    Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
    Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).
     
    Output
    For each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1.
     
    Sample Input
     3 100
    10 20
    45 89
    5 40
     
    3 100
    10 20
    45 90
    5 40
     
    3 100
    10 20
    45 84
    5 40
    Sample Output
    3 2 -1
    分析:
    (1)因为数据比较少,每一组最多有是个数据,可以用全排列暴力计算找出最小的组合
    (2)这里借助于stl算法库中的next_permutation()全排列算法。暴力把所有的排列方法计算一遍。然后找出最小的组合方法。
    #include <iostream>
    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    int main()
    {
        int o[10],a[10],m[10];
        int n,M,count,min;
    
        while(cin>>n>>M)
        {
            for(int i=0;i<n;i++)
            cin>>a[i]>>m[i];
            for(int i=0;i<n;i++)
            o[i] = i;
            min = 100;
            do{
                count = 0;
                int tem = M;
                for(int i=0;i<n;i++)
                {
                    count++;
                    if(tem<=m[o[i]])
                    tem -= a[o[i]]*2;
                    else tem -= a[o[i]];
                    if(tem<=0)
                    {
                        if(count<min)
                        min = count;
                        break;
                    }
                }
            }while(next_permutation(o,o+n));
            if(min==100)printf("-1\n");
            else printf("%d\n",min);
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/newpanderking/p/2713240.html
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