[ACM] hdu 4403 A very hard Aoshu problem (DFS暴搜数字）

A very hard Aoshu problem

Problem Description

Aoshu is very popular among primary school students. It is mathematics, but much harder than ordinary mathematics for primary school students. Teacher Liu is an Aoshu teacher. He just comes out with a problem to test his students:

Given a serial of digits, you must put a '=' and none or some '+' between these digits and make an equation. Please find out how many equations you can get. For example, if the digits serial is "1212", you can get 2 equations, they are "12=12" and "1+2=1+2". Please note that the digits only include 1 to 9, and every '+' must have a digit on its left side and right side. For example, "+12=12", and "1++1=2" are illegal. Please note that "1+11=12" and "11+1=12" are different equations.

 

Input

There are several test cases. Each test case is a digit serial in a line. The length of a serial is at least 2 and no more than 15. The input ends with a line of "END".

 

Output

For each test case , output a integer in a line, indicating the number of equations you can get.

 

Sample Input

1212123456661235END

 

Sample Output

220

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

解题思路：

对DFS还是很头疼，搞不懂它是怎么运行的，有时候看着别人写的DFS看着看着就晕了，还得多做做这类的题目，上次网赛就有一道DFS暴搜，不会啊。。。。

本题的题意为给定一个由数字组成的字符串，向里面加入一个等号，和任意个加号，使得等式成立，问一共有多少种方案。

思路为枚举等号出现的位置，用equ表示，则1<=equ<=len-1,  也就是把等号加在第equ个数字后面，紧跟着第equ个数字。然后先DFS等号左边所有可能出现的情况，对应每个左边所出现的和lsum，再DFS等号右边所有可能出现的情况rsum，如果 rsum==lsum，那么有一种方案可行。

总结：通过DFS来搜出等号左边所有的状态，每一种状态压入栈中，然后从栈中取状态，再去DFS等号右边所有的状态，状态相等，则有一种方案可行。

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;char str[17];int val[17][17];int len;int ans;void cal(){    memset(val,0,sizeof(val));    for(int i=1;i<=len;i++)        for(int j=i;j<=len;j++)    {        for(int k=i;k<=j;k++)            val[i][j]=val[i][j]*10+(str[k]-'0');    }}void dfsr(int lsum,int pos,int rsum)//dfs等号后边可能出现的情况{    if(pos>len)    {        if(rsum==lsum)            ans++;        return;    }    for(int k=pos;k<=len;k++)//注意是<=，因为最后一位也属于等号后边        dfsr(lsum,k+1,rsum+val[pos][k]);//注意是k+1}void dfsl(int equ,int pos,int lsum)//dfs等号左边可能出现的情况，每种情况对应一个lsum（等号左边的和）{    if(pos>equ)//当等号左边的数字的位置大于等号的位置时，搜索等号后边        dfsr(lsum,equ+1,0);    for(int k=pos;k<=equ;k++)//注意是<=,equ代表的是等号加在第equ个数字后面        dfsl(equ,k+1,lsum+val[pos][k]);}int main(){    while(cin>>(str+1)&&str[1]!='E')    {        ans=0;        len=strlen(str+1);        cal();        int equ;        for(equ=1;equ<=len-1;equ++)            dfsl(equ,1,0);        cout<<ans<<endl;    }    return 0;}