ZOJ 1204 Additive equations

先上原题

Additive equations
Time Limit: 10000 msMemory Limit: 32768 KB
We all understand that an integer set is a collection of distinct integers. Now the question is: given an integer set, can you find all its addtive equations? To explain what an additive equation is, let’s look at the following examples:
1+2=3 is an additive equation of the set {1,2,3}, since all the numbers that are summed up in the left-hand-side of the equation, namely 1 and 2, belong to the same set as their sum 3 does. We consider 1+2=3 and 2+1=3 the same equation, and will always output the numbers on the left-hand-side of the equation in ascending order. Therefore in this example, it is claimed that the set {1,2,3} has an unique additive equation 1+2=3.
It is not guaranteed that any integer set has its only additive equation. For example, the set {1,2,5} has no addtive equation and the set {1,2,3,5,6} has more than one additive equations such as 1+2=3, 1+2+3=6, etc. When the number of integers in a set gets large, it will eventually become impossible to find all the additive equations from the top of our minds – unless you are John von Neumann maybe. So we need you to program the computer to solve this problem.

Input

The input data consists of several test cases.
The first line of the input will contain an integer N, which is the number of test cases.
Each test case will first contain an integer M (1<=M<=30), which is the number of integers in the set, and then is followed by M distinct positive integers in the same line.

Output

For each test case, you are supposed to output all the additive equations of the set. These equations will be sorted according to their lengths first( i.e, the number of integer being summed), and then the equations with the same length will be sorted according to the numbers from left to right, just like the sample output shows. When there is no such equation, simply output “Can’t find any equations.” in a line. Print a blank line after each test case.

Sample Input
3
3 1 2 3
3 1 2 5
6 1 2 3 5 4 6
Output for the Sample Input
1+2=3

Can’t find any equations.

1+2=3
1+3=4
1+4=5
1+5=6
2+3=5
2+4=6
1+2+3=6

思路

1、首先，我要解释下我是怎么找出这道题的解法是深度搜索的。

假设我输入了一组数据：4 1 2 3 4（意味着有4个数，分别为1 2 3 4）
那么我要怎么找出所有的 Addtive Equations 呢？
我是通过一棵树来寻找的：

在这张图中，深度越深，代表找到的这个 Addtive Equations 等号左边的数越多。比如第四层就是代表1+2+3+4 。
我们要按照这张图的方法去遍历输入的每一组数据，显然，要用到的方法是深搜。

代码

（代码注释得很清晰，我就不多解释了）

#include <iostream>
#include <algorithm>
using namespace std;int set[100];//保存输入的每一组数据
bool visit[100];//用于记录有无访问过的点，用于输出 Addtive Equations 等号左边的数
bool has;//判断当前组数据有无 Addtive Equation//初始化set和visit
void init() {for (int i = 0; i < 100; ++i) {set[i] = 0;visit[i] = 0;}
}void display(int setnum,int sum) {int i = 0;//找到第一个访问过的数while ( !visit[i] && i<setnum ) ++i;cout << set[i];i = i + 1;//找到后续访问过的数while (i < setnum) {if (visit[i]) {cout << "+" << set[i];}++i;}cout << "=" << sum<<endl;
}void DFS(int sum,int num,int pos,int depth) {//sum保存以访问点的和，num代表这组数据中有多少个数，pos指出当前访问点在set中的下标位置，depth记录着与目标深度的距离//如果当前访问的数就是一组数据最后的一个数，那么后面肯定是没有数等于它和前面访问过的数的和if (pos >= num) return;//找到一个Addtive Equation，同时是符合当前需要寻找的深度if (!depth) {for (int i = pos; i < num; ++i) {if (set[i] == sum && visit[i] == 0) {//输出这一个 Addtive Equationdisplay(num, sum);has = 1;break;}}}for (int k = pos; k < num; ++k) {//剪枝，如果当前访问点和之前访问点的和大于这一组数据的最后一个数（最后一个数也是最大一个数），//那么之前访问点和后面将要访问的点的和肯定也大于最后一个数//不剪枝AC不了，会超时if (sum + set[k] > set[num - 1]) break;//对当前访问点进行深度搜索visit[k] = 1;DFS(sum + set[k], num, k + 1,depth-1);//不走当前访问点visit[k] = 0;}
}int main()
{int setNum; int lengths;//lengths是搜索的深度，代表着Addtive Equations等号左边的整数数量cin >> setNum;while (setNum) {int num;init();lengths = 2;//Addtive Equations等号左边最少也要两个数has = 0;cin >> num;for (int i = 0; i < num; ++i) {cin >> set[i];}//对左边整数集进行排序sort(set, set + num);while (lengths <= num) {DFS(0, num, 0, lengths);++lengths;}if (!has) cout << "Can't find any equations." << endl;cout << endl;setNum = setNum - 1;}system("pause");return 0;
}

踩坑

第一次我没有进行剪枝，在oj平台上怎么也AC不了（显示超时）。以后深搜如果超时了，要进行剪枝。

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