ZOJ2734-Exchange Cards-回溯法
2016-10-01 22:36
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题目大意:一个人有若干张卡片。每张卡片都有各自的价值,各个卡片价值可以一样,但还是属于不同的卡片,现在给你n,问用那叠卡片组成价值为n有多少种方法;
题目解析:可以暴力回溯,从第i张卡片之后开始枚举递归,注意要恢复状态;
AC代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
int sum,card[1010],way,n,m;
void fun(int t)
{
int i;
if(sum==n)
{
way++;
return ;
}
for(i=t;i<=n;i++)
{
if(card[i]&&i+sum<=n)
{
card[i]--;
sum+=i;
fun(i);
card[i]++;
sum-=i;
}
}
}
int main()
{
int i,a,b;
while(scanf("%d%d",&n,&m)!=EOF)
{
way=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
card[a]=b;
}
sum=0;
fun(1);
printf("%d\n",way);
}
return 0;
}
题目解析:可以暴力回溯,从第i张卡片之后开始枚举递归,注意要恢复状态;
AC代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
int sum,card[1010],way,n,m;
void fun(int t)
{
int i;
if(sum==n)
{
way++;
return ;
}
for(i=t;i<=n;i++)
{
if(card[i]&&i+sum<=n)
{
card[i]--;
sum+=i;
fun(i);
card[i]++;
sum-=i;
}
}
}
int main()
{
int i,a,b;
while(scanf("%d%d",&n,&m)!=EOF)
{
way=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
card[a]=b;
}
sum=0;
fun(1);
printf("%d\n",way);
}
return 0;
}
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