【思维】Kenken Race

题目描述

There are N squares arranged in a row, numbered 1,2,...,N from left to right. You are given a string S of length N consisting of . and #. If the i-th character of S is #, Square i contains a rock; if the i-th character of S is ., Square i is empty.
In the beginning, Snuke stands on Square A, and Fnuke stands on Square B.
You can repeat the following operation any number of times:
Choose Snuke or Fnuke, and make him jump one or two squares to the right. The destination must be one of the squares, and it must not contain a rock or the other person.
You want to repeat this operation so that Snuke will stand on Square C and Fnuke will stand on Square D.
Determine whether this is possible.

Constraints
4≤N≤200000
S is a string of length N consisting of . and #.
1≤A,B,C,D≤N
Square A, B, C and D do not contain a rock.
A, B, C and D are all different.
A<B
A<C
B<D

输入

Input is given from Standard Input in the following format:

N A B C D
S

 

输出

Print Yes if the objective is achievable, and No if it is not.

样例输入

7 1 3 6 7
.#..#..

样例输出

Yes

提示

The objective is achievable by, for example, moving the two persons as follows. (A and B represent Snuke and Fnuke, respectively.)

A#B.#..
A#.B#..
.#AB#..
.#A.#B.
.#.A#B.
.#.A#.B
.#..#AB

 

---

 

【题解】：

　　题目没有给定A,C与B,D之间的关系，所以我们分类讨论，

　　如果C==D，则无解，

　　如果两个区间不重叠，那么我们只要单独判断每一个区间内是否合法。

　　如果两个区间是相交的，那么我们需要判断一下中间是否有三个空位让A跳过去。先判断B是否能到D，如果不能，则无解，如果可以，还需要在路途中A，C路上是否有三个阻碍物，不然无法跳出去。

 

#include<bits/stdc++.h>
using namespace std ;
const int N = 2e5+100;
char s[N];
int main (){
    int n,A,B,C,D;
    scanf("%d%d%d%d%d",&n,&A,&B,&C,&D);
    scanf("%s",s+1);
    if( C == D ){
        printf("No\n");
    }else if( C < D ){
        int flag = 1 ;
        // if( s[C] == '#' || s[D] == '#' ) flag = 0 ;
        for ( int i = A ; i < C ; i++ ){
            if ( s[i] == '#' && s[i+1] == '#' ){
                flag = 0;
                break ;
            }
        }
        for ( int i = B ; i < D ; i++ ){
            if ( s[i] == '#' && s[i+1] == '#' ){
                flag = 0;
                break ;
            }
        }
        if( flag ){
            printf("Yes\n");
        }else{
            printf("No\n");
        }
    }else{
        int flag = 1 ;
        for ( int i = B ; i < D ; i++ ){
            if ( s[i] == '#' && s[i+1] == '#' ){
                flag = 0;
                break ;
            }
        }
        for ( int i = A ; i < C ; i++ ){
            if ( 
            (s[i] == '#' && s[i+1] == '#') ||
            (s[i] == '#' && i+1 == D ) ||
            (s[i+1] == '#' && i == D ) 
            ){
                flag = 0;
                break ;
            }
        }
         
        int f = 0 ;
        for ( int i = B  ; i <= D-1 ; i++ ){
            if ( s[i-1] == '.' && s[i] == '.' && s[i+1] == '.' ){
                f = 1 ;
                break;
            }
        }
        if( flag || f ){
            printf("Yes\n");
        }else{
            printf("No\n");
        }
    }
    return 0 ;
}