hdu4403暴力搜索

题意:
     给你一个数字串,让你在里面添加一个=和若干个+,使等式成立.


思路:

     lmax最大是15,直接暴搜,无压力,关键是判重,要在答案的时候判重,一开始在进队列之前判的,各种wa,哎!后来才发现如果在之前判断就不能得到当前什么符号都不加,而下一个有符号了,判重我用的是map,随意什么只要别爆内存就行,水搜索竟然调了2个小时,丢脸啊...

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

using namespace std;

typedef struct
{
   int mk[16];
   char str[16];
   int nowid ,deng;
}NODE;

NODE xin ,tou;
int nn;
map<__int64 ,__int64>mark;

bool MK(NODE aa)
{
   __int64 sum = 0;
   for(int i = 0 ;i <= nn ;i ++)
   sum = sum * 10 + aa.mk[i];
   if(mark[sum]) return 1;
   mark[sum] = 1;
   return 0;
}

bool ok(NODE tou)
{
   if(tou.deng)
   {
         int l ,r;
         l = r = 0;
         int sum = 0;
         int mki;
         for(int i = 0 ;tou.mk[i] != 2 ;i ++)
         {
            if(tou.mk[i] == 1)
            {
               l += sum;
               sum = tou.str[i] - 48;
            }
            else
            {
               sum = sum * 10 + tou.str[i] - 48;
            }
            mki = i;
         }
         l += sum;
         sum = 0;
         for(int i = mki + 1;i <= nn ;i ++)
         {
            if(tou.mk[i] == 1)
            {
               r += sum;
               sum = tou.str[i] - 48;
            }
            else
            {
               sum = sum * 10 + tou.str[i] - 48;
            }
         }
         r += sum;
    return l == r && !MK(tou);    
    }
    return 0;
}

int BFS()
{
   memset(xin.mk ,0 ,sizeof(xin.mk));
   nn = strlen(xin.str) - 1;
   xin.nowid = 0;
   xin.deng = 0;
   queue<NODE>q;
   q.push(xin);
   int ans = 0;
   mark.clear();
   while(!q.empty())
   {
      tou = q.front();
      q.pop();
      if(ok(tou))
      ans ++;
      if(tou.nowid == nn) continue;      
      if(!tou.deng)// =
      {
         xin = tou;
         xin.deng = 1;
         xin.nowid = tou.nowid + 1;
         xin.mk[xin.nowid] = 2;
         q.push(xin);  
      }
         
      //+
      xin = tou;
      xin.deng = tou.deng;
      xin.nowid = tou.nowid + 1;
      xin.mk[xin.nowid] = 1;
      q.push(xin);
      
      
      //
      xin = tou;
      xin.deng = tou.deng;
      xin.nowid = tou.nowid + 1;
      xin.mk[xin.nowid] = 0;
      q.push(xin);
   } 
   return ans;
}

int main ()
{
   while(~scanf("%s" ,xin.str) && strcmp("END" ,xin.str))
   {
      printf("%d
" ,BFS());
   }
   return 0;
}