Coder-Strike 2014 - Finals (online edition, Div. 1)

CF 420A  A. Start Up

题目链接：

http://codeforces.com/problemset/problem/420/A

题目意思：

给一个字符串A，通过镜面反射后得到A‘，判断A和A’是否完全一样。

解题思路：

水题。分析知有些字母通过镜面反射后和原来一样。

代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 33int hav[Maxn];char save[110000];void init(){    memset(hav,0,sizeof(hav));    hav[1]=1,hav['H'-'A'+1]=1;    hav['I'-'A'+1]=1;    hav['M'-'A'+1]=1;    hav['O'-'A'+1]=1;    hav['T'-'A'+1]=1;    hav['U'-'A'+1]=1;    hav['V'-'A'+1]=1;    hav['W'-'A'+1]=1;    hav['X'-'A'+1]=1;    hav['Y'-'A'+1]=1;}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%s",save+1))   {       int len=strlen(save+1);       int ans=1;       int p=1;       init();       //printf("%d %d\n",p,len);       while(save[p]==save[len]&&p<=len)       {           if(!hav[save[p]-'A'+1])           {               ans=0;               break;           }           p++;           len--;       }       if(p<=len)            ans=0;       if(ans)            printf("YES\n");       else            printf("NO\n");   }   return 0;}

CF 420B B. Online Meeting

题目链接：

http://codeforces.com/problemset/problem/420/B

题目意思：

有n个人开会，已知按时间顺序的m个进出记录（部分开会），求哪些人可能一直在开会。

解题思路：
抽象模拟。

可以保存一个当前时间的统治者me.

（无）+（有）

（有）-（无）

详细解释请见代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 110000int ans[Maxn],add[Maxn];int n,m,a,cnt,me;char temp[3];int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%d%d",&n,&m))   {       for(int i=1;i<=n;i++)            ans[i]=1,add[i]=0;;       cnt=0;       me=0;       for(int i=1;i<=m;i++)       {           scanf("%s%d",temp,&a);           if(*temp=='+')           {                if(!me||(me==a)) //更新统治者 注意要分别 +1 -1 +1的情况，此时统治者还是1                    me=a;                else                    ans[a]=0;   //和前面的通知不一样 不行 比如+1 +2那么2不行                add[a]=1;  //保存出现                if(a!=me&&!add[me]) //-1 +2 此时的1不行                    ans[me]=0;                cnt++;           }           else   //'-'           {               if(!me) //更新统治者                    me=a;              if(a!=me&&add[a]==0) //+1 +2 -3 此时1为统治者               {                   ans[me]=0;                   me=a;               }               if(add[a]) //抵消               {                   add[a]=0;                   cnt--;               }               if(cnt) //还有剩余力量 +1 +2 -2 此时的2不行 +1 +2 -1此时的1也不行，只要有+的不一样就不行                    ans[a]=0;           }       }       int tt=0;       bool fi=true;       for(int i=1;i<=n;i++)            if(ans[i])                tt++;       printf("%d\n",tt);       for(int i=1;i<=n;i++)       {           if(ans[i])           {               if(!fi) //第一个出现                    printf(" ");               printf("%d",i);               if(fi)                   fi=false;           }       }       printf("\n");   }   return 0;}

Codeforces 420C - Bug in Code题目链接：http://codeforces.com/problemset/problem/420/C解题思路：先预处理出当选中第i个人时，会使num[i]个人满意。注意当(i,j)是某人同时讨厌的时候，如果此时选中(i,j)只会使一个人满意。所以先出于处理出这种情况，直接判断出选中(i,j)时是否可行，如果不行提前减掉。然后再排序，枚举选中的第一个人，直接二分出第二个人的位置。代码:

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 310000int num[Maxn],n,p;map<pair<int,int>,int>myp;int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%d%d",&n,&p))   {       memset(num,0,sizeof(num));       myp.clear();       for(int i=1;i<=n;i++)       {           int a,b;           scanf("%d%d",&a,&b);           if(a>b)                swap(a,b);           myp[make_pair(a,b)]++;           num[a]++;           num[b]++;       }       map<pair<int,int>,int>::iterator it=myp.begin();       ll ans=0;       for(;it!=myp.end();it++)       {           int ta=it->first.first,tb=it->first.second;           if(num[ta]+num[tb]>=p&&num[ta]+num[tb]-it->second<p)                ans--;  //说明选中ta和tb不可行       }       sort(num+1,num+n+1);       for(int i=1;i<n;i++)       {          /* if(!num[i])                continue;*/           if(num[i]>=p)           {               ans+=n-i;               continue;           }           int temp=p-num[i];           if(temp>num[n]) //二分出第二个人的位置                continue;           int pos=lower_bound(num+i+1,num+n+1,temp)-num;           ans+=n-pos+1;           //printf("i:%d pos:%d\n",i,pos);       }       printf("%I64d\n",ans);   }   return 0;}