BNUOJ-26476 Doorman 贪心
题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=26476
题意：给一个字符序列，比如MWMMW，每次可以取前面两个中的一个，取出来后，取出来的那个个数加一，要求使得两个字符的个数不超过n，求最多能取多少个。。
贪心就可以了。
1 //STATUS:C++_AC_36MS_1480KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 //#include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef long long LL; 34 typedef unsigned long long ULL; 35 //const 36 const int N=110; 37 const int INF=0x3f3f3f3f; 38 const int MOD=1e9+7,STA=8000010; 39 //const LL LNF=1LL<<60; 40 const double EPS=1e-8; 41 const double OO=1e15; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 int n,vis[N],ans; 59 char s[N]; 60 int main(){ 61 // freopen("in.txt","r",stdin); 62 int i,j,l,r; 63 while(~scanf("%d",&n)) 64 { 65 scanf("%s",s); 66 67 mem(vis,0); 68 int len=strlen(s); 69 int cntm,cntw; 70 cntm=cntw=ans=0; 71 for(i=0;i<len;i++){ 72 l=r=-1; 73 for(j=0;j<len;j++){ 74 if(!vis[j]){ 75 if(l==-1)l=j; 76 else {r=j;break;} 77 } 78 } 79 if(cntm<=cntw){ 80 if(s[l]=='M')cntm++,vis[l]=1; 81 else if(s[r]=='M')cntm++,vis[r]=1; 82 else cntw++,vis[l]=1; 83 } 84 else { 85 if(s[l]=='W')cntw++,vis[l]=1; 86 else if(s[r]=='W')cntw++,vis[r]=1; 87 else cntm++,vis[l]=1; 88 } 89 if(abs(cntm-cntw)>n)break; 90 ans++; 91 } 92 93 printf("%d\n",ans); 94 } 95 return 0; 96 }