CodeForces 719B Anatoly and Cockroaches (水题贪心)

题意：给定一个序列，让你用最少的操作把它变成交替的，操作有两种，任意交换两种，再就是把一种变成另一种。

析：贪心，策略是分别从br开始和rb开始然后取最优，先交换，交换是最优的，不行再变色。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn];

int solve(char ch, char sh){
    int cntr = 0, cntb = 0, ans = 0;
    for(int i = 0; i < n; ++i){
        if(i & 1){
            if(s[i] == ch) continue;
            if(s[i] == sh){
                if(cntr) --cntr;
                else ++ans, ++cntb;
            }
        }
        else{
            if(s[i] == sh)  continue;
            if(s[i] == ch){
                if(cntb)  --cntb;
                else ++ans, ++cntr;
            }
        }
    }
    return ans;
}

int main(){
    while(scanf("%d", &n) == 1){
        scanf("%s", s);

        int ans = Min(solve(‘b‘, ‘r‘), solve(‘r‘, ‘b‘));
        printf("%d\n", ans);
    }
    return 0;
}