#include <iostream>
using namespace std;
const int MAX_NUM = 32;
const int MAX_MINUTE = 20 * 12;
int d[MAX_NUM], t[MAX_NUM], time[MAX_NUM], tt[MAX_NUM];
int f[MAX_NUM][MAX_MINUTE];
int n, h;
int max(int a, int b){return a > b ? a : b;}
void solve()
{
int i, k, min5, fish, max_fish = 0;
memset(time, 0, sizeof(time));
time[1] = h;
//枚举所经过的湖的个数k,每种情况减去所需步行的时间,剩下的就是钓鱼的时间了,
//之后就只需要在所剩的时间内使钓鱼数最大,因此每次选择最多鱼的湖点,
//最后构造起时间段。
for (k = 1; k <= n; ++k){
memset(tt, 0, sizeof(tt)); //tt是指每个湖当前已经进行到哪个时刻
min5 = h, fish = 0;
for (i = 2; i <= k; ++i) min5 -= t[i]; //去掉路上的时间
if (min5 <= 0) continue; //如果全部花在路上,则跳出
//贪心策略:分配每一个5分钟,找各个湖中当前鱼数最大的时间段
while (min5--){
int m = 0, p = 1;
for (i = 1; i <= k; ++i){
if (f[i][tt[i]+1] > m){ //找各个湖中当前鱼数最大的时间段
m = f[i][tt[i]+1];
p = i;
}
}
tt[p]++;
fish += m;
}
if (fish > max_fish){ //如果大于最大值,则更新时间段
max_fish = fish;
for (i = 1; i <= k; ++i){
time[i] = tt[i];
}
for (i = k+1; i <= n; ++i){
time[i] = 0;
}
}
}
for (i = 1; i < n; ++i){
printf("%d, ", time[i]*5);
}
printf("%d\n", time[i]*5);
printf("Number of fish expected: %d\n\n", max_fish);
}
int main()
{
int i, j;
while (scanf("%d", &n) && n != 0){
scanf("%d", &h);
h *= 12;
for (i = 1; i <= n; ++i){
scanf("%d", &f[i][1]);
}
for (i = 1; i <= n; ++i){
scanf("%d", &d[i]);
}
for (i = 2; i <= n; ++i){
scanf("%d", &t[i]);
}
//计算每个湖,不同时间段中的鱼数,单位是5分钟
//f[i][j]表示第i个湖,第j个5分钟能钓到的鱼数
for (i = 1; i <= n; ++i){
for (j = 2; j <= h; ++j){
f[i][j] = max(f[i][j-1]-d[i], 0);
}
}
solve();
}
return 0;
}
POJ 1042(枚举+贪心)
最新推荐文章于 2019-04-19 08:38:47 发布