Gone Fishing 贪心+枚举

//关键点：求出在第一个湖耗时最长的，所以如果有多余的时间则尽量都耗在第一个湖边
//贪心+枚举
#include <stdio.h>
#include <string.h>
int initialLakeFishes[26], fishReduce[26], moveTime[26];
int timeSpend[26], eachFish[26], totalFish[26], copyInitialLake[26];

struct Node{
	int lake, fish;
} maxLake;

int findMaxLake(int n){
	maxLake.lake = maxLake.fish = 0;
	int i;
	for(i = 0; i <= n; ++i){
		if(copyInitialLake[i] > maxLake.fish)
			maxLake.fish = copyInitialLake[i], maxLake.lake = i;		
	}
	return maxLake.fish;
}

void mergeSum(int n){
	int sum = 0;
	for(int i = 0; i <= n; ++i)
		sum += eachFish[i];
	totalFish[n] = sum;
}

int findMaxTotal(int n){
	int i, max = 0, j = 0;
	for(i = 0; i <= n; ++i)
		if(totalFish[i] > max)
			max = totalFish[i], j = i;
	return j;
}

int main(){
	int lakeNumbers, hours;
	int hoursToMinutes, copy_hoursToMinutes, i, n;
	while(scanf("%d", &lakeNumbers), lakeNumbers){
		scanf("%d", &hours);
		hoursToMinutes = hours * 60;
		copy_hoursToMinutes = hoursToMinutes;
		for(i = 0; i != lakeNumbers; ++i)
			scanf("%d", &initialLakeFishes[i]);
		for(i = 0; i != lakeNumbers; ++i)
			scanf("%d", &fishReduce[i]);
		for(i = 0; i != lakeNumbers - 1; ++i)
			scanf("%d", &moveTime[i]);
		//从第一个湖开始枚举,假设小明最远到达第n个湖
		//则小明可以在这n个湖之间“瞬移”。		
		for(n = 0; n != lakeNumbers; ++n){
			memcpy(copyInitialLake, initialLakeFishes, sizeof(initialLakeFishes));
			for(i = 0; i < n; ++i){
				hoursToMinutes -= moveTime[i] * 5;
			}
			while(hoursToMinutes >= 5){
				if(findMaxLake(n) == 0)
					break;
				timeSpend[maxLake.lake] += 5;
				hoursToMinutes -= 5;
				eachFish[maxLake.lake] += maxLake.fish;
				copyInitialLake[maxLake.lake] -= fishReduce[maxLake.lake];
			}
			mergeSum(n);
			memset(eachFish, 0, sizeof(eachFish));
			memset(timeSpend, 0, sizeof(timeSpend));
			hoursToMinutes = copy_hoursToMinutes;
		}
		
		memcpy(copyInitialLake, initialLakeFishes, sizeof(initialLakeFishes));
		n = findMaxTotal(lakeNumbers - 1);
		for(i = 0; i < n; ++i){
			hoursToMinutes -= moveTime[i] * 5;
		}
		
		while(hoursToMinutes >= 5){
			if(findMaxLake(n) == 0)
				break;
			timeSpend[maxLake.lake] += 5;
			hoursToMinutes -= 5;
			eachFish[maxLake.lake] += maxLake.fish;
			copyInitialLake[maxLake.lake] -= fishReduce[maxLake.lake];
		}
		while(hoursToMinutes >= 5){
			timeSpend[0] += 5;
			hoursToMinutes -= 5;
		}
			
		printf("%d", timeSpend[0]);
		for(i = 1; i != lakeNumbers; ++i)
			printf(", %d", timeSpend[i]);		
		printf("\nNumber of fish expected: %d\n\n", totalFish[n]); 
		memset(totalFish, 0, sizeof(totalFish));
		memset(eachFish, 0, sizeof(eachFish));
		memset(timeSpend, 0, sizeof(timeSpend));
	}
	return 0;
}