poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)

Sum It Up

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 1

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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

浙江工业大学第四届大学生程序设计竞赛

做练习的时候，纠结死啦！

当时我一直在想怎样判重，自己YY了一个hash函数 WA了。。。

网上的一种解法不需要判重，剪枝就可以了。按照那个思路我重新敲了一下。

code1: (dfs+剪枝)

poj:

Accepted

388K

0MS

G++

930B

#include <stdio.h>
#include <string.h>

int a[15];
int p[15];
int vis[15];
int t, n, flag;

void dfs(int k, int sum) {
    int i;
    if(k>n || sum<0) return ;
    if(sum==0) {
        flag = 1;
        for(i=0; i<k-1; i++)
            printf("%d+",p[i]);
        printf("%d\n",p[i]);
        return ;
    }
    for(i=k; i<n; i++)
        if(!vis[i]) {
            if(sum-a[i]<0||(k>0&&a[i]>p[k-1])) continue;
            vis[i] = 1;
            p[k] = a[i];
            dfs(k+1,sum-a[i]);
            vis[i] = 0;
            while(i+1<n&&a[i]==a[i+1]) i++; //搜索完毕后，若下一个搜索的数仍与当前相同，则寻找下一个不同的数进行搜索。｛去重｝
        }
}
int main() {
    int i;
    while(scanf("%d%d",&t,&n),t+n) {
        for(i=0; i<n; i++) scanf("%d",&a[i]);
        i = 0;
        while(i<n&&a[i]>t) i++;
        printf("Sums of %d:\n",t);
        flag = 0;
        memset(vis,0,sizeof(vis));
        dfs(i,t);
        if(!flag) printf("NONE\n");
    }
    return 0;
}

code2:(用 set 去重：在POJ和ZOJ上提交全挂，不过hdu上能AC，呃呃呃~) 

HDU:

Accepted

1258

0MS

340K

1286 B

G++

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <vector>
#include <string>
#include <set>
using namespace std;
#define N 20
int t, n;
int a[N];
int list[N];
bool vis[N];
set<string> s;
bool flag;

void dfs(int k, int sum) {
    int i;
    if(k>=n || sum<0) return;
    if(sum==0) {
        string str;
        for(i=0; i<k; i++) {
            str +=(list[i]/10) +'0';
            str +=(list[i]%10) +'0';
        }
        if(s.find(str)==s.end()) {
            s.insert(str);
            flag = 1;
            for(i=0; i<k-1; i++)
                printf("%d+",list[i]);
            printf("%d\n",list[i]);
        }
        return ;
    }
    for(i=k; i<n; i++)
        if(!vis[i]&&(k==0||a[i]<=list[k-1])) {
            vis[i] = 1;
            list[k] = a[i];
            dfs(k+1,sum-a[i]);
            vis[i] = 0;
        }
}
int main() {
    int i;
    while(scanf("%d%d",&t,&n),t+n) {
        for(i=0; i<n; i++) {
            scanf("%d",&a[i]);
        }
        memset(vis,0,sizeof(vis));
        printf("Sums of %d:\n",t);
        flag = false;
        s.clear();
        dfs(0,t);
        if(!flag) printf("NONE\n");
    }
    return 0;
}