uva 757 Gone Fishing

 

 

John is going on a fishing trip. He has h hours available ( ), and there are n lakes in the area ( ) 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 , the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted t_i ( ). 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 f_i ( ), 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 d_i ( ). If the number of fish expected to be caught in an interval is less than or equal to d_i, 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 f_i ( ), then a line of n integers d_i ( ), and finally, a line of n - 1 integers t_i ( ). 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.

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

 

 

 

題目大意：

每組樣例包括5各部分：

1）魚塘數 n

2）總時間 h（小時）

3）n 個魚塘，每個魚塘中初始魚的產量（每5分鐘可以釣上的魚的數量）

4）n 個魚塘，每個魚塘魚的損耗速度（每在該魚塘釣魚5分鐘，魚的產量減少的數量）

5）n 個魚塘之間的路程的耗時（單位為5分鐘）

要求在總時間 h 內可以釣上的最多的魚的數量，和在每一個魚塘停留的時間。當到達下一個魚塘時，不能返回上一個魚塘。

解題思路：先要對魚塘之間路程的耗時進行處理，要將其轉變為從第一個魚塘到第 i 個魚塘所需的時間。

枚舉只在前 i 個魚塘釣魚的 n 種情況，這樣 只在前 i 個魚塘釣魚的釣魚時間 = 總時間 - 第一個魚塘到第 i 個魚塘所需時間，然後在前 i 個魚塘中找出當前魚產量最多的魚塘，在此釣魚，然後重復之前的過程，直到前 i 個魚塘無魚可釣或者時間耗盡。這樣，n 種情況中的最大值，就是可以釣起的最多魚的數量。

 

 

 

#include
#include
#include
#include
using namespace std;
int n, h, f[250], d[250], t[250], ans, rec[250], recA[250], temp[250];
int find(int x) {
	int num = 0, id;
	for (int i = 0; i <= x; i++) {
		if (temp[i] > num) {
			num = temp[i];
			id = i;
		}	
	}
	if (num) {
		return id;
	}
	else return -1;
}
int main() {
	int flag = 0;
	while (scanf("%d", &n) == 1, n) {
		if (flag) printf("\n");
		flag = 1;
		int H;
		scanf("%d", &H);
		h = H * 12;
		for (int i = 0; i < n; i++) {
			scanf("%d", &f[i]);
		}
		for (int i = 0; i < n; i++) {
			scanf("%d", &d[i]);
		}
		t[0] = 0;
		for (int i = 1; i < n; i++) {
			scanf("%d", &t[i]);
			t[i] += t[i - 1];
		}
		memset(recA, 0, sizeof(recA));
		ans = 0;	
		for (int i = 0; i < n; i++) { //枚舉n種情況：從第一個魚塘到第i個魚塘
			int time = h - t[i], sum = 0;
			if (time <= 0) break;
			memcpy(temp, f, sizeof(f));
			memset(rec, 0, sizeof(rec));
			while (time) {
				int id = find(i); //find找出前i個魚塘中魚最多的魚塘
				if (id < 0) break;
				sum += temp[id];
				temp[id] -=	d[id];	
				time--;
				rec[id]++;
			}
			if (time > 0) { //當前i個魚塘都無魚可釣，且還有剩余時間，留在第一個魚塘
				rec[0] += time;
			}
			if (sum > ans) {
				ans = sum;
				memcpy(recA, rec, sizeof(rec));
			}
		}
		if (ans) {
			printf("%d", recA[0] * 5);
		}
		else printf("%d", h * 5); //當無魚可釣的時候也要停留在第一個魚塘，樣例會卡這裡
		for (int i = 1; i < n; i++) {
			printf(", %d", recA[i] * 5);
		}
		printf("\nNumber of fish expected: %d\n", ans);
	}	
	return 0;
}