白昼夢 / Daydream（AtCoder-2158）

Problem Description

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S=T by performing the following operation an arbitrary number of times:

Append one of the following at the end of T: dream, dreamer, erase and eraser.

Constraints

• 1≦|S|≦105
• S consists of lowercase English letters.

Input

The input is given from Standard Input in the following format:

S

Output

If it is possible to obtain S=T, print YES. Otherwise, print NO.

Example

Sample Input 1

erasedream

Sample Output 1

YES
Append erase and dream at the end of T in this order, to obtain S=T.

Sample Input 2

dreameraser

Sample Output 2

YES
Append dream and eraser at the end of T in this order, to obtain S=T.

Sample Input 3

dreamerer

Sample Output 3

NO

题意：给一个字符串 S 与一个空串 T，现在可以在 T 的末尾添加以下任何一项：dream、dreamer、erase、eraser，问能否添加任意次后使得 S=T

思路：模拟即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 100000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;
char str[N];
char Dream[5]= {'d','r','e','a','m'};
char Erase[5]= {'e','r','a','s','e'};
int main() {
    scanf("%s",str);
    int len=strlen(str);

    bool flag=true;
    for(int i=0; i<len; i++) {
        if(str[i]=='d') {
            int j=i;
            for(; j<=i+4; j++) {
                if(j>=len||str[j]!=Dream[j-i]) {//是否为dream或dreamer
                    flag=false;
                    break;
                }
            }
            if(j==i+5) {
                if(str[j]=='e'&&str[j+1]=='r'&&str[j+2]!='a')//dreamer，指针前移
                    i=j+1;
                else//dream，指针后移
                    i=j-1;
            }
        }
        else if(str[i]=='e') {
            int j=i;
            for(; j<=i+4; j++) {
                if(j>=len||str[j]!=Erase[j-i]) { //是否为erase或eraser
                    flag=false;
                    break;
                }
            }
            if(j==i+5) {
                if(str[j]=='r')//eraser，指针前移
                    i=j;
                else//erase，指针后移
                    i=j-1;
            }
        }
        else
            flag=false;

        if(!flag)
            break;
    }

    if(flag)
        printf("YES\n");
    else
        printf("NO\n");

    return 0;
}