POJ 1042 Gone Fishing（模拟+贪心）

题目地址：http://poj.org/problem?id=1042

思路：模拟+贪心

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct node
{
    int first;
    int reduce;
}a[30];
int b[30];
int time[30];
int pre[30][30];
int c[30];
int main()
{
    int n,h;
    while(scanf("%d",&n) && n)
    {
        scanf("%d",&h);
        h *= 12;
        memset(pre,0,sizeof(pre));
        memset(c,0,sizeof(c));
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i].first);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i].reduce);
        time[1] = 0;
        for(int i=2; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            time[i] = time[i-1] + x;
        }
        for(int i=1; i<=n; i++)
        {
            int total = h - time[i];
            for(int j=1; j<=i; j++)
                b[j] = a[j].first;
            for(int j=1; j<=total; j++)
            {
                int max1 = b[1];
                int tt = 1;
                for(int k=1; k<=i; k++)
                {
                    if(b[k] > max1)
                    {
                        max1 = b[k];
                        tt = k;
                    }
                }
                b[tt] -= a[tt].reduce;
                if(b[tt] < 0)
                    b[tt] = 0;
                c[i] += max1;
                pre[i][tt]++;
            }
        }
        int max1 = -1;
        int l;
        for(int i=1; i<=n; i++)
        {
            if(c[i] > max1)
            {
                max1 = c[i];
                l = i;
            }
        }
        for(int i=1; i<=n-1; i++)
            printf("%d, ",pre[l][i]*5);
        printf("%d\n",pre[l][n]*5);
        printf("Number of fish expected: %d\n\n",max1);
    }
    return 0;
}