BOI2007 Mokia
分治套树状数组，板子题
1 #include<bits/stdc++.h> 2 #define FOR(i,a,b) for(int i=a,ie=b;i<=ie;++i) 3 #define For(i,a,b) for(int i=a,ie=b;i<ie;++i) 4 using namespace std; 5 int read(){ 6 int x=0,f=1;char c=getchar(); 7 for(;!isdigit(c);c=getchar())if(c=='-')f=-1; 8 for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48); 9 return x*f; 10 } 11 const int maxn=4e5+10; 12 typedef long long ll; 13 14 int ans[maxn],a[maxn]; 15 struct oper{ 16 int x1,x2,y,f,id;//divide the y 17 oper(){}; 18 oper(int x1,int x2,int y,int f,int id): 19 x1(x1),x2(x2),y(y),f(f),id(id){}; 20 }q[maxn]; 21 22 bool cmp(oper a,oper b){ 23 if(a.y!=b.y)return a.y<b.y; 24 return a.id<b.id; 25 } 26 27 struct TAS{ 28 int tr[maxn]; 29 int lb(int x){return x&-x;} 30 void add(int x,int v){ 31 for(;x<maxn;x+=lb(x))tr[x]+=v; 32 } 33 int sum(int x){ 34 int ret=0; 35 for(;x>0;x-=lb(x))ret+=tr[x]; 36 return ret; 37 } 38 }T; 39 40 void solve(int l,int r){ 41 if(l==r)return; 42 int mid=l+r>>1,j=0; 43 solve(l,mid);solve(mid+1,r); 44 vector<oper> s1,s2; 45 s1.clear(),s2.clear(); 46 FOR(i,l,mid)if(!q[i].id)s1.push_back(q[i]); 47 FOR(i,mid+1,r)if(q[i].id)s2.push_back(q[i]); 48 sort(s1.begin(),s1.end(),cmp); 49 sort(s2.begin(),s2.end(),cmp); 50 For(i,0,s2.size()){ 51 for(int je=s1.size();j<je;++j) 52 if(s1[j].y<=s2[i].y) 53 T.add(s1[j].x1,s1[j].f); 54 else break; 55 ans[s2[i].id]+=s2[i].f*T.sum(s2[i].x2); 56 ans[s2[i].id]-=s2[i].f*T.sum(s2[i].x1-1); 57 } 58 For(i,0,j)T.add(s1[i].x1,-s1[i].f); 59 } 60 61 int main(){ 62 int op=read(),n=read(),m=0,id_cnt=0; 63 for(int x1,x2,y1,y2,v;(op=read())!=3;){ 64 if(op&1){ 65 x1=read();y1=read();v=read(); 66 q[++m]=oper(x1,x1,y1,v,0);a[m]=x1; 67 } 68 else{ 69 x1=read();y1=read(); 70 x2=read();y2=read(); 71 q[++m]=oper(x1,x2,y2,1,++id_cnt);a[m]=x2; 72 q[++m]=oper(x1,x2,y1-1,-1,id_cnt);a[m]=x1-1; 73 } 74 } 75 sort(a+1,a+m+1); 76 int cnt=unique(a+1,a+m+1)-(a+1); 77 for(int i=1;i<=m;++i){ 78 if(!q[i].id) 79 q[i].x1=lower_bound(a+1,a+cnt+1,q[i].x1)-a; 80 else{ 81 q[i].x1=lower_bound(a+1,a+cnt+1,q[i].x1-1)-a+1; 82 q[i].x2=lower_bound(a+1,a+cnt+1,q[i].x2)-a; 83 } 84 } 85 solve(1,m); 86 for(int i=1;i<=id_cnt;++i)printf("%d\n",ans[i]); 87 return 0; 88 }
