HDU1258————DFS

Sum It Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others) Total Submission(s): 9199 Accepted Submission(s):
4792

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

#include<bits/stdc++.h>
using namespace std;

int a[13];
int n, t;
int vis[15];//标记数组
int flag;//标记是否有答案
void DFS(int i, int sum)
{

    if(sum == t&&n+1 == i)
    {
        flag = 1;
        int f = 0;
        for(int k = 1; k <= n; k++)
        {
            if(vis[k])
            {
                if(f)printf("+%d", a[k]);
                else printf("%d", a[k]), f= 1;
            }
        }
        printf("\n");
        return ;
    }
    else if(n+1 == i)return;
    vis[i] = 1;
    DFS(i+1, sum+a[i]);
    vis[i] = 0;
    //回溯到层选择不选时 重复的↓
    while(i+1 <=n&&a[i] == a[i+1])i++;//如果这个数不选，要把重复取掉， 某地就是为了防止出现重复答案
    DFS(i+1, sum);
}
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>t>>n&&(t||n))
    {
        flag = 0;

        for(int i = 1; i <= n; i++)
            cin>>a[i];


        printf("Sums of %d:\n", t);
        DFS(1, 0);
        if(!flag)
        {
            printf("NONE\n");
        }

    }

    return 0;
}

暴力点的做法
dfs记录所有答案再查重

#include<bits/stdc++.h>
using namespace std;

int a[13];
int n, t;
int vis[15];
int ans[100000][17];
int j = 0;
void DFS(int i, int sum)
{

    if(sum == t&&n+1 == i)
    {
        j++;
        for(int k = 1; k <= n; k++)
        {
            if(vis[k])
            {
                ans[j][0]++;
                ans[j][ans[j][0]] = a[k];
            }
        }
        return ;
    }
    else if(n+1 == i)return;
    vis[i] = 1;
    DFS(i+1, sum+a[i]);
    vis[i] = 0;
    DFS(i+1, sum);
}
int main()
{
    while(cin>>t>>n&&(t||n))
    {
        j = 0;

        for(int i = 1; i <= n; i++)
            cin>>a[i];


        memset(vis, 0, sizeof(vis));
        memset(ans, 0, sizeof(ans));
        DFS(1, 0);

        printf("Sums of %d:\n", t);
        if(j == 0)
        {
            printf("NONE\n");
        }
        else
        {

            for(int i  = 1; i <= j; i++)
            {
                if(ans[i][0] == -1)continue;
                for(int ii = i+1; ii <= j; ii++)
                {
                    if(ans[ii][0] == -1)continue;
                    int com = 0;
                    if(ans[i][0] == ans[ii][0])
                    {


                        for(int k = 1; k<= ans[i][0]; k++)
                        {
                            if(ans[i][k]!=ans[ii][k])
                            {
                                com = 1;
                                break;
                            }
                        }
                        if(!com)
                        {
                            ans[ii][0]= -1;
                        }
                    }
                }
            }

            for(int i = 1; i <= j; i++)
            {
                if(ans[i][0] == -1)continue;
                for(int k = 1; k <= ans[i][0] ; k++)
                {
                    if(k == 1)printf("%d", ans[i][k]);
                    else printf("+%d", ans[i][k]);
                }
                printf("\n");
            }

        }
    }
    return 0;
}