HDU 1258Sum It Up(暴力dfs，記住相同的狀態只保留一個)

果7發表於2013-10-17

Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3300    Accepted Submission(s): 1665


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





 



Source



浙江工業大學第四屆大學生程式設計競賽



 








題目大意：給你一個數的和sum，最大1000，然後是n（n<=12）個<=100的正數數，a[0],a[1]...a[n-1]，選取這裡面的數，使得和為sum，開始準備暴力列舉。。不用那麼麻煩，可以直接暴搜，不過遇到同一個狀態則保留一個，在搜尋的時候每一個點訪問之後，這個值在這個相同的位置不能再次被訪問。因為這點一樣後面的情況肯定一樣了。具體實現見程式碼。





題目地址：Sum It Up





AC程式碼：




#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[15],sum,n;
int ans[15];
int flag,tt;

bool cmp(int a,int b)
{
    return a>b?1:0;
}

void dfs(int t,int s)
{
    int i;
    if(s>sum) return;
    if(s==sum)
    {
        cout<<ans[0];
        for(i=1;i<tt;i++)
            cout<<"+"<<ans[i];
        cout<<endl;
        flag=1;   //說明找到了一個
    }

    int last=-1;
    for(i=t+1;i<n;i++)
    {
        if(a[i]!=last)  //相應的同一個位置的節點不允許相同
        {
            last=a[i];
            ans[tt++]=a[i];
            dfs(i,s+a[i]);
            tt--;
        }
    }
}

int main()
{
    int i;
    while(scanf("%d%d",&sum,&n))
    {
        flag=0;
        if(!sum&&!n) break;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n,cmp);
        printf("Sums of %d:\n",sum);
        int last=-1;
        for(i=0;i<n;i++)
        {
            if(a[i]!=last)  //每次起點不同
            {
                last=a[i];
                tt=0;
                ans[tt++]=a[i];
                dfs(i,a[i]);
            }
        }
        if(!flag)
            puts("NONE");
    }
    return 0;
}

//0MS





            




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