tzc1909 Gone Fishing【枚举+贪心】

 Gone Fishing

时间限制(普通/Java):2000MS/20000MS     运行内存限制:65536KByte
总提交: 13            测试通过: 4

描述

John is going on a fishing trip. He has h hours available (1 <= h <= 16), and there are n lakes in the area (2 <= n <= 25) all reachable along a single, one-way road. John starts at lake 1, but he can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each i = 1,...,n - 1, the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti (0 < ti <=192). For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4. To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi( fi >= 0 ), is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di (di >= 0). If the number of fish expected to be caught in an interval is less than or equal to di , there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at the lakes to affect the number of fish he expects to catch.
Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.

输入

You will be given a number of cases in the input. Each case starts with a line containing n. This is followed by a line containing h. Next, there is a line of n integers specifying fi (1 <= i <=n), then a line of n integers di (1 <=i <=n), and finally, a line of n - 1 integers ti (1 <=i <=n - 1). Input is terminated by a case in which n = 0.

输出

For each test case, print the number of minutes spent at each lake, separated by commas, for the plan achieving the maximum number of fish expected to be caught (you should print the entire plan on one line even if it exceeds 80 characters). This is followed by a line containing the number of fish expected.
If multiple plans exist, choose the one that spends as long as possible at lake 1, even if no fish are expected to be caught in some intervals. If there is still a tie, choose the one that spends as long as possible at lake 2, and so on. Insert a blank line between cases.

样例输入

 

2
1
10 1
2 5
2
4
4
10 15 20 17
0 3 4 3
1 2 3
4
4
10 15 50 30
0 3 4 3
1 2 3
0

 

样例输出

 

45, 5
Number of fish expected: 31

240, 0, 0, 0
Number of fish expected: 480

115, 10, 50, 35
Number of fish expected: 724

分析：

意思的话自己读吧  关键是意思读清楚 这里的从1开始往i湖走去  走到i 再回去的话就不花费走路
的时间了。简单说就是走路的时间只花费一次。
这点题目读到了就可以了。没有读到的话很郁闷的说。。。。  然后的话枚举到i个湖的时候
 h时间减去路上花费的时间 剩余的时间就是可以在i
个湖里瞬间移动 不花费时间的了。所以只要去最大鱼数量的湖里吊就行了。
得到best[i] .枚举到湖n个 再 max =max{best[i]} i=1,2,3,...n
OK 解决了看参考code:
 #include <cstdio>
#include<iostream>
#include<cstdlib>
#include<queue>
using namespace std;
#define maxn 27
#define inf 0x7fffffff
#define eps 1e-11
typedef struct nod
{
    int w;
    int f;
    bool operator<(const nod &b)const
    {
        if(b.f==f)
        {
            return b.w<w;
        }
        else
        {
            return b.f>f;
        }
    }
}node;
priority_queue<node> v;
int main()
{
    int h,n,i,j,th,max,sum,flag=0;
    int di[maxn],ti[maxn],stp[maxn],tstp[maxn];
    node fi[maxn];
    while(scanf("%d",&n),n)
    {
        if(flag)printf("/n");flag=1;
        max=-1;
        memset(tstp,0,sizeof(tstp));
        
        scanf("%d",&h);
        h=h*12;
        for(i=0;i<n;i++)
        {    scanf("%d",&fi[i].f);
            fi[i].w=i;
        }
        for(i=0;i<n;i++)
            scanf("%d",&di[i]);
        for(i=0;i<n-1;i++)
        {scanf("%d",&ti[i]);}
        
        for(i=0;i<n;i++)
        {
            while(v.size()){v.pop();}
            for(j=0;j<=i;j++)
                v.push(fi[j]);
            
            sum=0;
            for(th=1;th<=h;th++)
            {
                
                node tt;
                tt=v.top();
                v.pop();
                sum+=tt.f;
                
                tstp[tt.w]+=1;
                
                tt.f-=di[tt.w];
                if(tt.f<0){tt.f=0;}
                
                v.push(tt);
            }
                
            if(max<sum){max=sum;memcpy(stp,tstp,sizeof(tstp));}
            
                memset(tstp,0,sizeof(tstp));
            
            h-=ti[i];
        }
        
        

        for(i=0;i<n-1;i++)
        {
            printf("%d, ",stp[i]*5);
        }
        printf("%d/n",stp[i]*5);
        printf("Number of fish expected: %d/n",max);
        
        
    }
    return 0;
}