题目链接:http://codeforces.com/problemset/problem/962/B
There are nn consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of aa student-programmers and bb student-athletes. Determine the largest number of students from all a+ba+b students, which you can put in the railway carriage so that:
- no student-programmer is sitting next to the student-programmer;
- and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
The first line contain three integers nn, aa and bb (1≤n≤2?1051≤n≤2?105, 0≤a,b≤2?1050≤a,b≤2?105, a+b>0a+b>0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length nn, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
5 1 1 *...*
2
6 2 3 *...*.
4
11 3 10 .*....**.*.
7
3 2 3 ***
0
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
题意:
题目大致意思是说,给出你一行座位,然后有程序员和运行员两种人,要求两种人不能相邻地坐着,给出你n个座位,a个程序员和b个运动员,然后还有座位上现在的情况,一个字符串,*代表有人,.代表没人可以坐下来,然后问你最多可以在这一列座位下做多少人?
标签给的是贪心,自己想了一下,想取几次min,然后判断一下,但是交了几次改了几次,都没有过到,后来索性删了,重新换个思路。
对于每个座位,首先,要求可以做下来最多的人,那么第一个座位就可以让当前的人最多的程序员或者运动员坐,偶数个连坐的时候没有问题,奇数的时候,就会多出来一个,这样就可以避免再特判了,然后每次坐了一种人之后,如果连坐,下一个座位应该是另外一种人,所以可以定义一个标记记录一下。在不能做的情况下,每次去重置一下标记,让最多那类人首先坐下。
代码如下:
#include <bits/stdc++.h>using namespace std;int n,a,b;
char str[202000];
//vector <int > v;
int main() {while(cin >> n >> a >> b) {scanf("%s",str);
// v.clear();int n1 = a, m = b;bool flag;if(a > b)flag = 1;else flag = 0;for(int i = 0; i < n; i++) {if(str[i] == '.') {if(a > 0 && flag) {a--;flag = 0;continue;} else if(b > 0 && !flag) {b--;flag = 1;continue;}}if(a > b) flag = 1;else flag = 0;}int ans = (n1 - a) + (m - b);cout << ans << endl;}return 0;
}