POJ2892 Tunnel Warfare

题解:

树状数组+二分裸题。

对于二分时的细节问题，要仔细分析再确定.

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define LL long long
#define CLR(x) memset(x,0,sizeof x)
#define MC(x,y) memcpy(x,y,sizeof(x))
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=50010;
const int mod=1e9+7;
const int INF=1e9;

int n,m,c[maxn];
int pre[maxn],x,cnt;
char op;

int lowbit(int x){return x&-x;}

void add(int x,int v){
    while(x<=n){
        c[x]+=v;
        x+=lowbit(x);
    }
}

int sum(int x){
    int num=0;
    while(x){
        num+=c[x];
        x-=lowbit(x);
    }
    return num;
}

int L_Bin(int l,int r){
   int R=r,m,goal=r;
   while(l<r){
         m=(l+r)>>1;
         if(sum(R)-sum(m)==R-m){
              goal=m+1;
              r=m;
         }
         else l=m+1;
   }
   return goal;
}

int R_Bin(int l,int r){
   int L=l,m,goal=l;
   while(l<=r){
         m=(l+r)>>1;
         if(sum(m)-sum(L)==m-L){
              goal=m;
              l=m+1;
         }
         else r=m-1;
   }
   return goal;
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        CLR(c);
        cnt=0;
        for(int i=1;i<=n;i++) add(i,1);
        for(int i=1;i<=m;i++){
           scanf(" %c",&op);
           if(op==‘D‘){
                 scanf("%d",&x);
                 add(x,-1);
                 pre[cnt++]=x;
           }
           if(op==‘R‘) add(pre[--cnt],1);
           if(op==‘Q‘){
                scanf("%d",&x);
                if(sum(x)-sum(x-1)==0){
                    printf("0\n");
                    continue;
                }
                int L=L_Bin(0,x);
                int R=R_Bin(x,n);
                //cout<<L<<" "<<R<<endl;
                printf("%d\n",R-L+1);
           }
        }
    }
    return 0;
}