hdu 1258 深度优先搜索+步长+序列和+输出+虚拟节点+文件操作

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Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5473    Accepted Submission(s): 2862


Problem Description

Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that equal 4: 4,3+1,2+2, and 2+1+1.(A number
can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

 

Input

The input will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x1,...,xn. If n=0 it signals the end of the input; otherwise, t will be a positive integer
less than 1000, n will be an integer between 1 and 12(inclusive), and x1,...,xn will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.

 

Output

For each test case, first output a line containing 'Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line 'NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated
in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number
must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distince; the same sum connot appear twice.

 

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

 
分析：深度优先搜索各节点，建立一个虚拟节点0，将n棵树的森林连成一棵树，最重要的是 去重 ，方法是，对一个节点搜索完毕后，若下一个节点与当前节点相同，则跳过直至找到一个不同节点。

代码：

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<stack>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
# define maxn 1005
int a[maxn] ;
int chose[maxn] ;
int visited[maxn] ;
int t , n ,flag;
void dfs(int v ,int sum , int step ){
//printf("chose vex = %d ,\n" , v ) ;
if(sum > t)
return ;
if(sum == t){
flag = 1 ;
for(int i = 1 ; i<= step ; i++){
if(i == 1)
printf("%d" , chose[i]) ;
else
printf("+%d" , chose[i]) ;
}
cout<< endl;
return ;
}
for(int i =1 ; i <= n ; i++){
if(!visited[i] && i> v && sum + a[i] <= t){
visited[i] = 1 ;
chose[step + 1 ] = a[i] ;
dfs(i , sum + a[i] , step + 1) ;
//printf("return i= %d\n", i) ;
visited[i] = 0 ;
while(i+1 <= n && a[i+1] == a[i]) //搜索完毕后，若下一个搜索数与当前相同，则跳过直至不同
i++ ;
}
}
}
int main(){
// FILE * fp ;
// fp = fopen("in.txt" , "r") ;
while(scanf("%d%d" , &t,&n) != EOF){

if(n == 0){
break ;
}
flag = 0 ;
for(int i =1 ; i<= n ;i++)
scanf("%d" , &a[i]) ;
//fscanf(fp , "%d" , &a[i]) ;
memset(visited , 0 , sizeof(visited)) ;
printf("Sums of %d:\n" , t ) ;

dfs(0 , 0 , 0) ; //引入虚拟节点0,将n个森林连成一棵树
if(!flag)
printf("NONE\n") ;
}
//fclose(fp) ;
return 0 ;
}