Codeforces Round #433 (Div. 1, based on Olympiad of Metropolises) C. Boredom

需要算所有跟查询矩形相交的矩形（两个点组成的）
只要把所有矩形的数量减去不相交的矩形就好了

显然不相交的矩形是有在查询矩形的上、下、左、右的两个点组成
但是会有重复的部分即在左上左下右上右下的部分
所以再把答案加回这部分

利用主席树查询[l,r]上[d,u]的数有多少个

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  200000+10;const int maxm = maxn*21+10;//Pretests passedint in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}int root[maxn];struct Seg{    int cnt,lch,rch;}seg[maxm];int tot;void update(int &rt,int l,int r,int x){    int last =  rt;    seg[rt=++tot] = seg[last];    ++seg[rt].cnt;    if(l==r)return ;    int mid = (l+r)>>1;    if(x<=mid)update(seg[rt].lch,l,mid,x);    else update(seg[rt].rch,mid+1,r,x);}int query(int rt,int L,int R,int l,int r){    if(l<=L&&R<=r)return seg[rt].cnt;    int m = (L+R)>>1;    if(r<=m)return query(seg[rt].lch,L,m,l,r);    if(m<l)return query(seg[rt].rch,m+1,R,l,r);    return query(seg[rt].lch,L,m,l,r) + query(seg[rt].rch,m+1,R,l,r);}int n;int query(int xl,int xr,int yl,int yr){    return query(root[xr],1,n,yl,yr) - query(root[xl-1],1,n,yl,yr);}ll cal(int x){    return (1LL*(x-1)*x)>>1;}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int q;    sdd(n,q);    r1(i,n)    {        int x;        sd(x);        root[i] = root[i-1];        update(root[i],1,n,x);    }    while(q--)    {        int x1,y1,x2,y2;        sdd(x1,y1),sdd(x2,y2);        ll ans = cal(n);        ans -= cal(x1-1) + cal(n-x2) + cal(y1-1) +cal(n-y2);        int lu = 1<x1&&1<y1? query(1,x1-1,1,y1-1) : 0 ;        int ld = 1<x1&&y2<n? query(1,x1-1,y2+1,n) : 0 ;        int ru = x2<n&&y1>1? query(x2+1,n,1,y1-1) : 0 ;        int rd = x2<n&&y2<n? query(x2+1,n,y2+1,n) : 0 ;        ans += cal(lu) + cal(ld) + cal(ru) + cal(rd);        lansn();    }    return 0;}