URAL 1427. SMS(DP+单调队列)

题目链接

我用的比较传统的办法。。。单调队列优化了一下，写的有点搓，不管怎样过了。。。两个单调队列，存两个东西，预处理一个标记数组存。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <map>
 #include <ctime>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define INF 1000000
 char str[];
 int dp[];
 int pre[];
 int que1[];
 int que2[];
 int judge(char s)
 {
     if(s <= 'Z'&&s >= 'A')
         return ;
     else if(s <= 'z'&&s >= 'a')
         return ;
     else if(s == ' ')
         return ;
     else
         return ;
 }
 int main()
 {
     int n,m,i,j,len,str1,end1,str2,end2;
     scanf("%d%d%*c",&n,&m);
     gets(str);
     len = strlen(str);
     for(i = ; i < len; i ++)
     {
         if(judge(str[i]))
         {
             for(j = i; j < len; j ++)
             {
                 if(judge(str[j]))
                 pre[j] = i;
                 else
                 {
                     i = j;
                     break;
                 }
             }
             if(j == len) break;
         }
     }
     for(i = ;i <= len;i ++)
     dp[i] = INF;
     str1 = ;end1 = ;
     str2 = ;end2 = ;
     que1[] = que2[] = ;
     for(i = ;i <= len;i ++)
     {
         dp[i] = dp[que1[str1]] + ;
         if(judge(str[i-]))
         {
             if(str2 < end2)
             dp[i] = min(dp[i],dp[que2[str2]]+);
         }
         else
         {
             str2 = end2 = ;
         }
         if(i == len) continue;
         while(str1 < end1&&dp[i] <= dp[que1[end1-]])
         end1 --;
         que1[end1++] = i;
         while(str1 < end1&&i - que1[str1] >= n)
         str1 ++;
         if(judge(str[i]))
         {
             while(str2 < end2&&dp[i] <= dp[que2[end2-]])
             end2 --;
             que2[end2++] = i;
             while(str2 < end2&&que2[str2] < pre[i])
             str2 ++;
             while(i - que2[str2] >= m&&str2 < end2)
             str2 ++;
         }
     }
     printf("%d\n",dp[len]);
     return ;
 }