ABC047&ARC063.1D Reversi

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#ABC047&amp;ARC063 #1D Reversi

本文介绍了一款名为1D Reversi的游戏及其算法解决方案。游戏的目标是通过放置黑白棋子使所有棋子颜色一致。文章详细解释了游戏规则，并提供了一个具体的算法实现案例，该算法能够计算出达到目标所需的最少新棋子数。

问题 E: 1D Reversi

时间限制: 1 Sec 内存限制: 128 MB
提交: 332 解决: 139
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题目描述

Two foxes Jiro and Saburo are playing a game called 1D Reversi. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.
In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1≤i≤|S|) stone from the left. If the i-th character in S is B, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is W, it means that the color of the corresponding stone is white.
Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Constraints
1≤|S|≤105
Each character in S is B or W.

输入

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

输出

Print the minimum number of new stones that Jiro needs to place for his purpose.

样例输入

BBBWW

样例输出

提示

By placing a new black stone to the right end of the row of stones, all white stones will become black. Also, by placing a new white stone to the left end of the row of stones, all black stones will become white.
In either way, Jiro's purpose can be achieved by placing one stone.

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#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    string s;
    while(cin>>s){
        int len=s.size();
        int cnt=0,cnt1=0;
        for(int i=0;i<len-1;i++){
            if(s[i]==s[i+1]) cnt++;
            if(s[i]!=s[i+1]) {
                cnt1++;
            }
        }
        if(cnt==(len-1)) cout<<0<<endl;
        else cout<<cnt1<<endl;
    }
}