Gone Fishing
Time Limit: 2000MS | Memory Limit: 32768K | |
Total Submissions: 34616 | Accepted: 10727 |
Description
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.
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.
Input
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.
Output
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.
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.
Sample Input
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
Sample Output
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
//贪心
#include
#include
#include
#include
using namespace std;
struct node{
int w,i;
node(){}
node(int x,int y):w(x),i(y){}
bool operator<(node t)const{
if(w==t.w) return i>t.i;
return w>n&&n){
int fi[30],di[30],t[30]={0},x,ans=-1;
cin>>h;
h*=12;
for(int i=1;i<=n;i++)
cin>>fi[i];
for(int i=1;i<=n;i++)
cin>>di[i];
for(int i=2;i<=n;i++)
{
cin>>x;
t[i]=t[i-1]+x;
}
int a[30];
for(int i=1;i<=n;i++){//枚举每个点
int temp[30]={0};
priority_queue q;
for(int j=1;j<=i;j++)
q.push(node(fi[j],j));
int hh=h-t[i],cnt=0;
while((hh--)>0){
node k=q.top();
q.pop();
cnt+=k.w;
k.w-=di[k.i];
temp[k.i]++;
if(k.w<0) k.w=0;
q.push(k);
}
if(ans
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