[树状数组] hdu 5372 Segment Game

题目分析

    记录左端点、右端点两棵线段树。在添加某线段时，其完全所覆盖的前面的线段数量为：ans=tot−numl−r
    其中，tot为总线段数，numl为小于当前线段左端点的左端点数，r为大于当前线段右端点的右端点数，numl和r都是没有被覆盖的线段数量，用总数-不合题意数得到的就是合题数。（由于前面的线段长度短，因此无重复）。
    记小于等于当前线段右端点的右端点数为numr，则r=tot−numr
    故ans=numr−numl
    另外数据范围太大，需要离散化，由于需要的是左、右端点分别的对应大小关系，故只需分别离散化即可。

代码

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<map>#include<algorithm>#define lson l,mid,cur<<1#define rson mid+1,r,cur<<1|1typedef long long ll;using namespace std;const int maxn = 2e5+7;int maxb;int n;int btl[maxn], btr[maxn];inline int lowbit(int x){return x & -x;}void add(int bit[], int x, int val){for(int i = x; i <= maxb; i+=lowbit(i)) bit[i] += val;}int query(int bit[], int x){    int res = 0;    for(int i = x; i > 0; i -= lowbit(i)) res += bit[i];    return res;}int del[maxn],x[maxn],y[maxn],z[maxn];int ld,ly;int sub[maxn];void disc(int y[], const int &n){    for(int i=1;i<= n; i++) sub[i] = y[i];    sort(sub+1, sub+1 + n);    int *t = unique(sub+1, sub+1+n);    for(int i = 1; i <= n; i++)        y[i] = lower_bound(sub+1, t, y[i]) - sub;}int main(){    int cnt = 1;    while(cin>>n)    {        ld=ly=0;        printf("Case #%d:\n", cnt++);        for(int i = 1; i <= n; i++)        {            scanf("%d",&x[i]);            if(x[i])                scanf("%d", &del[++ld]);            else            {                scanf("%d", &y[++ly]);                z[ly] = y[ly] + ly;            }        }        maxb = ly;        for(int i  = 0; i <= maxb; i++)            btl[i] = btr[i] = 0;        disc(y,ly);        disc(z,ly);        int p1 = 0, p2 = 0;        for(int i = 1; i <= n; i++)        {            if(x[i])            {                add(btl, y[del[++p2]], -1);                add(btr, z[del[p2]], -1);            }            else            {                int l=y[++p1],r=z[p1];                printf("%d\n",query(btr,r)-query(btl,l-1));                add(btl,l,1);                add(btr,r,1);            }        }    }    return 0;}