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Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=2e5+10; const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}}; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,k,t,x[maxn],y[maxn],p[maxn],xcnt,ycnt,id,idx; ll ans; multiset<int>s; struct node { int x,low,high; bool operator<(const node&p)const { return x<p.x; } }a[maxn]; struct node1 { int x,y,type; bool operator<(const node1&p)const { return x<p.x; } }g[maxn]; int get(int x) { int sum=0; for(int i=x;i;i-=(i&(-i))) sum+=p[i]; return sum; } void add(int x,int y) { for(int i=x;i<=maxn-5;i+=(i&(-i))) p[i]+=y; } int main() { int i,j; scanf("%d",&t); while(t--) { xcnt=ycnt=id=idx=0; ans=0; memset(p,0,sizeof(p)); scanf("%d",&n); rep(i,1,n) { int b,c,d,e; scanf("%d%d%d%d",&b,&c,&d,&e); if(c>e)swap(c,e); if(b>d)swap(b,d); if(b==d) { x[++xcnt]=a[++id].x=b; y[++ycnt]=a[id].low=c; y[++ycnt]=a[id].high=e; } else { x[++xcnt]=b; x[++xcnt]=d; y[++ycnt]=c; g[++idx].x=b; g[idx].y=c; g[idx].type=0; g[++idx].x=d; g[idx].y=e; g[idx].type=1; } } sort(x+1,x+xcnt+1); sort(y+1,y+ycnt+1); sort(a+1,a+id+1); sort(g+1,g+idx+1); int num1=unique(x+1,x+xcnt+1)-x-1; int num2=unique(y+1,y+ycnt+1)-y-1; rep(i,1,id) { a[i].x=lower_bound(x+1,x+num1+1,a[i].x)-x; a[i].low=lower_bound(y+1,y+num2+1,a[i].low)-y; a[i].high=lower_bound(y+1,y+num2+1,a[i].high)-y; } rep(i,1,idx) { g[i].x=lower_bound(x+1,x+num1+1,g[i].x)-x; g[i].y=lower_bound(y+1,y+num2+1,g[i].y)-y; } int now=1; rep(i,1,id) { while(now<=idx&&g[now].x<=a[i].x) { if(g[now].type==0)add(g[now].y,1); else { if(g[now].x!=a[i].x) add(g[now].y,-1); else s.insert(g[now].y); } now++; } ans+=get(a[i].high)-get(a[i].low-1); for(int x:s)add(x,-1); s.clear(); } printf("%lld\n",ans); } //system("Pause"); return 0; }
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原文地址:http://www.cnblogs.com/dyzll/p/5790162.html