poj 1564 Sum It Up

最新推荐文章于 2021-03-20 09:42:49 发布

转载最新推荐文章于 2021-03-20 09:42:49 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/w0w0/archive/2012/05/08/2490044.html

此博客讨论了如何在给定整数列表中找出所有可能的组合，这些组合的和等于指定的目标值。通过使用深度优先搜索（DFS）算法，解决了一般问题并提供了优化的解决方案以避免重复计算。

Sum It Up

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4491		Accepted: 2264

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 x 1 , . . . , x n . 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 x 1 , . . . , x n 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 distinct; the same sum cannot 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

DFS的题，关键在于判重，这个方法还是看了网上别人的方法，很简单，但是很有效，自愧不如。。。。

#include<iostream>
#include<queue>
using namespace std;

int nums[15];
int t, n;
int ans[15];
int flag;

void dfs(int a, int j, int cur)
{
    if(a == 0)
    {
        cout<<ans[0];
        for(int i = 1; i<cur;++i)
        {
            cout<<"+"<<ans[i];
        }
        cout<<endl;
        flag = 1;
        return;
    }
    int pre = -1;
    for(int i = j; i < n; i++)
    {
        if(a>=nums[i] && nums[i]!=pre)
        {
            pre = nums[i];
            ans[cur] = nums[i];
            dfs(a-nums[i], i+1, cur+1);
        }
    }
}

int main()
{
    int i;
    freopen("e:\\data.txt", "r", stdin);   
    freopen("e:\\out.txt", "w", stdout); 
    while(cin>>t>>n && t && n)
    {
        flag = 0;
        for(i = 0; i < n; ++i)
        {
            cin>>nums[i];
        }
        cout<<"Sums of "<<t<<":"<<endl;
        
        dfs(t, 0, 0);
        
        if(flag == 0)
            cout<<"NONE"<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/w0w0/archive/2012/05/08/2490044.html