poj1564-Sum It Up(经典DFS)

给出一个n,k,再给出的n个数中，输出所有的可能使几个数的和等于k

Sample Input

4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0
Sample Output

Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25

明显的DFS，这个dfs方程让我纠结啊，递归的我头都大了，但是看下答案稍微来点灵感了,在这里dfs函数方程要有哪些参数？

首先要从当前数往后开始dfs，所以要有个参数是当前搜索的数组下标

其次，要判断和=t，所以还要有个保存当前的和的参数，在这里我用t减去当前和，所以当此参数等于0那么就是找到满足条件

最后要输出此序列，所以还要有个参数来标记当前要找的数在数组里的位置

找到三个参数后就好办多了，看代码吧

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

int n,t;

int num[101];

int r[101];

bool flag;

void dfs(int len,int k,int last)

{

int result;

if(last==0)

{

flag=false;

for(int i=0;i<len;i++)

if(i==0)printf("%d",r[i]);

else printf("+%d",r[i]);

printf("\n");

return ;

}

for(int i=k;i<n;i++)

{

if(i==k||num[i]!=num[i-1]&&last-num[i]>=0)// 去除重复的操作

{

r[len]=num[i];

dfs(len+1,i+1,last-num[i]);

}

int main()

{

while(scanf("%d%d",&t,&n)!=EOF)

{

if(t==n&&n==0)break;

flag=true;

for(int i=0;i<n;i++)

scanf("%d",&num[i]);

printf("Sums of %d:\n", t);

dfs(0,0,t);

if(flag)printf("NONE\n");

}

时间： 2024-10-06 20:11:06

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