hdu4403暴力搜索

来源：互联网发布：喝水提醒电脑软件编辑：程序博客网时间：2024/05/22 23:26

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\n" ,BFS());   }   return 0;}

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