pku 1042 Gone Fishing dp

#include <iostream>
using namespace std;
int s[26][17*12];
int f[26], d[26], t[26], ret[26];
int n, h;
void findPath(int i, int time)
{
if(i == 0) //此处要看我们dp时,从0开始,还是从1开始.
//(从0开始,就是if(i==0)return;
//从1开始,就是if(i==1){dosomething return;})
return;
int sum = 0;
int l = f[i];
for(int k = 0; k <= 12*h; ++k)
{
if(s[i-1][time-k-t[i-1]]+sum == s[i][time]) //回溯找路径的条件.
{
ret[i] = k*5;
return findPath(i-1, time-k-t[i-1]);
}
if(l > 0)
{
sum += l;
l -= d[i];
}
}
}
int main()
{
while(scanf("%d", &n) && n)
{
scanf("%d", &h);
for(int i = 1; i <= n; ++i)
scanf("%d", &f[i]);
for(int i = 1; i <= n; ++i)
scanf("%d", &d[i]);
for(int i = 1; i <= n-1; ++i)
scanf("%d", &t[i]);
memset(s, -1, sizeof(s));
s[0][0] = 0; //从0开始初始化
for(int i = 0; i <= n-1; ++i)
for(int j = 0; j <= 12*h; ++j)
{
if(s[i][j] != -1)
{
int sum = 0;
int l = f[i+1]; //由i推i+1的情况时,需要知道的i+1的情况.(Error: f[i])
for(int k = 0; k <= 12*h; ++k)
{
if(j+k+t[i] <= 12*h)
s[i+1][j+k+t[i]] = max(s[i+1][j+k+t[i]], s[i][j]+sum);
else
break;
if(l > 0)
{
sum += l;
l -= d[i+1]; //由i推i+1的情况时,需要知道的i+1的情况.(Error: d[i])
}
}
}
}
int max = -2;
int index = 1;
for(int i = 1; i <= n; ++i)
if(max < s[i][12*h])
{
max = s[i][12*h];
index = i;
}
memset(ret, 0, sizeof(ret)); //初始化,很重要!!
findPath(index, 12*h);
for(int i = 1; i <= n-1; ++i)
printf("%d, ", ret[i]);
printf("%d/n", ret
);
printf("Number of fish expected: %d/n/n", s[index][12*h]);
}
return 0;
}