zoj 2734 Exchange Cards 【DFS or 母函数】
As a basketball fan, Mike is also fond of collecting basketball player cards. But as a student, he can not always get the money to buy new cards, so sometimes he will exchange with his friends for cards he likes. Of course, different cards have different value, and Mike must use cards he owns to get the new one. For example, to get a card of value 10$, he can use two 5$ cards or three 3$ cards plus one 1$ card, depending on the kinds of cards he have and the number of each kind of card. And Sometimes he will involve unfortunately in a bad condition that he has not got the exact value of the card he is looking for (fans always exchange cards for equivalent value).
Here comes the problem, given the card value he plans to get and the cards he has, Mike wants to fix how many ways he can get it. So it's you task to write a program to figure it out.
Input
The problem consists of multiple test cases, terminated by EOF. There's a blank line between two inputs.
The first line of each test case gives n, the value of the card Mike plans to get and m, the number of different kinds of cards Mike has. n will be an integer number between 1 and 1000. m will be an integer number between 1 and 10.
The next m lines give the information of different kinds of cards Mike have. Each line contains two integers, val and num, representing the value of this kind of card, and the number of this kind of card Mike have.
Note: different kinds of cards will have different value, each val and num will be an integer greater than zero.
Output
For each test case, output in one line the number of different ways Mike could exchange for the card he wants. You can be sure that the output will fall into an integer value.
Output a blank line between two test cases.
Sample Input
5 2 2 1 3 1 10 5 10 2 7 2 5 3 2 2 1 5
Sample Output
1 7
题意：给您一个价值N，现在有M个卡片，每个卡片都有对应的价值与数目，问你有多少种方法用这些卡片构成价值N。
母函数基础题：0ms
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct rec { int val, num; }; rec TT[20]; int N, M; int c1[10100], c2[10100]; int main() { int t = 0; while(scanf("%d%d", &N, &M) != EOF) { if(t > 0) printf("\n"); for(int i = 1; i <= M; i++) scanf("%d%d", &TT[i].val, &TT[i].num); memset(c1, 0, sizeof(c1)); memset(c2, 0, sizeof(c2)); for(int i = 0, used = 0; used <= TT[1].num; i+=TT[1].val, used++) c1[i] = 1; for(int i = 2; i <= M; i++) { for(int j = 0; j <= N; j++) { for(int k = 0, used = 0; k + j <= N && used <= TT[i].num; k+=TT[i].val, used++) c2[k+j] += c1[j]; } for(int j = 0; j <= N; j++) { c1[j] = c2[j]; c2[j] = 0; } } printf("%d\n", c1[N]); t++; } return 0; }
DFS：0ms 拆分物品 + 剪枝去重
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int rec[10000010];
int ans;
int N, M;
int k;
void DFS(int len, int pos)
{
if(len == N)
{
ans++;
return ;
}
for(int i = pos; i < k; i++)
{
if(len + rec[i] <= N)
DFS(len + rec[i], i+1);
while(rec[i] == rec[i+1] && i < k)//去重
++i;
}
}
int main()
{
int t = 0;
while(scanf("%d%d", &N, &M) != EOF)
{
if(t > 0) printf("\n");
int x, y;
k = 0;
for(int i = 1; i <= M; i++)
{
scanf("%d%d", &x, &y);
while(y--)
rec[k++] = x;
}
ans = 0;
DFS(0, 0);
printf("%d\n", ans);
t++;
}
return 0;
}
第二种写法，看的别人的，感觉挺牛的。orz。。。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct rec
{
int val, num;
};
rec TT[20];
int ans;
int N, M;
void DFS(int len, int pos)
{
if(len == N)
{
ans++;
return ;
}
if(pos > M || len > N)
return ;
for(int i = 0; i <= TT[pos].num; i++)
DFS(len + TT[pos].val * i, pos+1);
}
int main()
{
while(scanf("%d%d", &N, &M) != EOF)
{
for(int i = 1; i <= M; i++)
scanf("%d%d", &TT[i].val, &TT[i].num);
ans = 0;
DFS(0, 1);
printf("%d\n", ans);
}
return 0;
}