hdu 3436 Queue-jumpers (经典离散化，树状数组实现lower_bound)

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3436

我的离散化是把 Top和Query的位置离散，然后把两两之间的位置合并成一个点，并记录个数num[i], 被Top， Query询问的位置也是一个点num[i] = 1; 小细节是左右两端的区间也要缩点。

#include<iostream>#include<cstdio>#include<cstring>#include<map>#include<algorithm>#include<cmath>#include<queue>#include<stack>using namespace std;#define M 300010#define inf 1<<30#define ll long long#define eps 1e-7#define bas 9#define mod 1000000000LLint sum[M];int bit[21];int N, n, m;void add( int i, int x ){    while( i <= N ){        sum[i] += x;        i += i&-i;    }}int query( int i ){    int ans = 0;    while( i ){        ans += sum[i];        i -= i&-i;    }    return ans;}char op[100010][6];int a[100010];int uni[M], cnt;int lisan[M], num[M], e, ma;int peo[M], pos[M];int find( int k ){    int ans = 0, pos = 0;    for( int i = 20; i >= 0; --i ){        pos += bit[i];        if( pos > N || ans + sum[pos] >= k ) pos -= bit[i];        else ans += sum[pos];    }    return pos+1;}int main(){    int T, t = 0;    bit[0] = 1;    for( int i = 1; i < 21; ++i ) bit[i] = bit[i-1]<<1;    scanf( "%d", &T );    while( T-- ){        scanf( "%d%d", &n, &m );        for( int i = 0; i < m; ++i )            scanf( "%s%d", op[i], &a[i] );        ma = cnt = 0;        for( int i = 0; i < m; ++i )            if(op[i][0]=='T'||op[i][0]=='Q')                uni[cnt++] = a[i];            else ma = max( ma, a[i] );        printf( "Case %d:\n", ++t );        sort( uni, uni+cnt );        if( ma > uni[cnt-1] ) uni[cnt++] = ma;        cnt = unique( uni, uni+cnt ) - uni;        e = 0;        if( uni[0] != 1 )            lisan[e] = 1, num[e++] = uni[0] - 1;        lisan[e] = uni[0], num[e++] =  1;        for( int i = 1; i < cnt; ++i ){            if( uni[i] - uni[i-1] > 1 )                lisan[e] = uni[i-1]+1, num[e++] = uni[i]-uni[i-1]-1;            lisan[e] = uni[i], num[e++] = 1;        }        N = e+m+3;        memset( sum, 0, sizeof(sum) );        int top = m;        for( int i = 0; i < e; ++i ){            add( top+i+1, num[i] );            pos[i] = top+i+1;     //记录人在树状数组的位置            peo[top+i+1] = i;      //记录树状数组所代表的人        }        for( int i = 0; i < m; ++i ){            if( op[i][0] == 'T' ){                int x = lower_bound( lisan, lisan+e, a[i] ) - lisan;                add( pos[x], -1 );                add( top, 1 );                peo[top] = peo[pos[x]];                pos[x] = top--;            }            else if( op[i][0] == 'Q' ){                int x = lower_bound( lisan, lisan+e, a[i] ) - lisan;                printf( "%d\n", query( pos[x] ) );            }            else{                int p = find( a[i] );                int s = query( p );                int k = peo[p];                //int x = lower_bound( lisan, lisan+e, k ) - lisan;                printf( "%d\n", lisan[k]+num[k]-s+a[i]-1 );            }        }    }}