bzoj1176

CDQ分治裸题

单点加询问子矩阵和

因为矩阵过大,所以树套树肯定是要跪的

那么我们考虑将子矩阵差分了，那么就是变成每次询问(0,0) 到(x,y)这个矩阵的和了

那么直接cdq分治就好了

/**************************************************************     Problem: 1176     User: syh0313     Language: C++     Result: Accepted     Time:13364 ms     Memory:158344 kb ****************************************************************/   #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <cmath> using namespace std; const int maxn=100010; int w,n,cnt,opt,x,xx,y,yy; long long s,ans[maxn],vv,a[50*maxn]; struct da{int op,x,y;long long v;int id;}q[50*maxn]; bool cmp(da aa,da bb){return aa.x<bb.x;} inline int lowbit(int x){return x&(-x);} void add(int x,long long k){while (x<=w+1) {a[x]+=k; x+=lowbit(x);}} long long qury(int x){long long sum=0;while (x) {sum+=a[x]; x-=lowbit(x);} return sum;} void cdq(int l,int r) {     if (l==r) return;     int mid=(l+r)>>1;     cdq(l,mid); sort(q+l,q+mid+1,cmp); cdq(mid+1,r); sort(q+mid+1,q+r+1,cmp);     int p1=l,p2=mid+1;     while (p2<=r)     {         while (p1<=mid && q[p1].x<=q[p2].x) {if (q[p1].op==1) add(q[p1].y,q[p1].v); p1++;}         if (q[p2].op==2) ans[q[p2].id]+=qury(q[p2].y)*q[p2].v;         p2++;     }     p1=l; p2=mid+1;     while (p2<=r)     {         while (p1<=mid && q[p1].x<=q[p2].x) {if (q[p1].op==1) add(q[p1].y,-q[p1].v); p1++;}         p2++;     } } int main() {     scanf("%lld%d",&s,&w);     while (scanf("%d",&opt) && opt!=3)     {         if (opt==1) {scanf("%d%d%lld",&x,&y,&vv); x++; y++; q[++n]={1,x,y,vv,0};}         else         {             scanf("%d%d%d%d",&x,&y,&xx,&yy); cnt++; x++; y++; xx++; yy++;             q[++n]={2,x-1,y-1,1,cnt}; q[++n]={2,xx,yy,1,cnt};             q[++n]={2,x-1,yy,-1,cnt}; q[++n]={2,xx,y-1,-1,cnt};         }     }     cdq(1,n); for (int i=1;i<=cnt;i++) printf("%lld\n",ans[i]+s); return 0; }