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.
輸入
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.
輸出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.
樣例輸入
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
樣例輸出
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
題意：John要釣魚，有 n 個魚塘，John有 t 小時的時間，每個魚塘中有 F[i] 條魚，每次釣魚這個魚塘就會減少 D[i] 條魚，從魚塘 i-1 到 i 需要花費 T[i] 分鐘，求出怎麼能夠得到最多的魚數，並求出在每個魚塘花費的時間。每次釣魚花費5分鐘，且給出的時間一定是5分鐘的倍數。
是一道貪心題目，因爲每次魚的減少情況只與在這個魚塘的釣魚次數有關，所以是貪心。我們枚舉前 i 個魚塘能夠得到的最大魚數，並且計算每個魚塘花費的時間，選出最大數量作爲答案。
我使用了優先隊列，這樣排序比較方便。
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
struct node
{
int no, fish, down, time;
friend bool operator < (const node &a, const node &b)
{
if (a.fish != b.fish)
return a.fish < b.fish;
return a.no > b.no;
}
};
node pool[30];
priority_queue<node> Q;
int h, n, maxsum;
int ans[30], time[30];
void greedy()
{
maxsum = -1;
for (int i = 0; i < n; ++i)
{
while (!Q.empty())
Q.pop();
for (int j = 0; j <= i; ++j)
Q.push(pool[j]);
int leave = h - pool[i].time;
int sum = 0;
memset(time, 0, sizeof(time));
while (leave > 0)
{
node tmp = Q.top();
Q.pop();
if (tmp.fish <= 0)
break;
sum += tmp.fish;
tmp.fish -= tmp.down;
time[tmp.no] += 5;
Q.push(tmp);
leave -= 5;
}
if (leave > 0)
time[0] += leave;
if (sum > maxsum)
{
maxsum = sum;
for (int j = 0; j < n; ++j)
ans[j] = time[j];
}
}
}
int main()
{
while (scanf("%d", &n) != EOF && n != 0)
{
scanf("%d", &h);
h *= 60;
for (int i = 0; i < n; ++i)
{
scanf("%d", &pool[i].fish);
pool[i].no = i;
}
for (int i = 0; i < n; ++i)
scanf("%d", &pool[i].down);
pool[0].time = 0;
for (int i = 1; i < n; ++i)
{
int t;
scanf("%d", &t);
pool[i].time = pool[i-1].time + 5*t;
}
greedy();
for (int i = 0; i < n; ++i)
{
printf("%d", ans[i]);
if (i < n-1)
printf(", ");
}
printf("\n");
printf("Number of fish expected: %d\n", maxsum);
}
return 0;
}