1176: [Balkan2007]Mokia
分析
三维偏序问题，CDQ分治论文题。
代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 inline int read() { 6 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 7 for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 8 } 9 10 int ans[10010],W; 11 struct Que{ 12 int ty,x,y,w,id; 13 bool operator < (const Que &a) const { 14 return x == a.x ? ty < a.ty : x < a.x; 15 } 16 }A[160000*4+10000+10],B[160000*4+10000+10]; 17 18 struct Bit{ 19 int sum[2000010]; 20 void update(int p,int v) { 21 for (; p<=W; p+=p&(-p)) sum[p] += v; 22 } 23 int query(int p) { 24 int ans = 0; 25 for (; p; p-=p&(-p)) ans += sum[p]; 26 return ans; 27 } 28 void clear(int p) { 29 for (; p<=W&&sum[p]; p+=p&(-p)) sum[p] = 0; 30 } 31 }bit; 32 33 void CDQ(int L,int R) { 34 if (L == R) return ; 35 int M = (L + R) >> 1; 36 CDQ(L,M); 37 CDQ(M+1,R); 38 int i = L, j = M + 1,k = L; 39 while (i <= M && j <= R) { 40 if (A[i] < A[j]) { 41 if (A[i].ty == 0) bit.update(A[i].y,A[i].w); 42 B[k++] = A[i++]; 43 } 44 else { 45 if (A[j].ty == 1) ans[A[j].id] += A[j].w * bit.query(A[j].y); 46 B[k++] = A[j++]; 47 } 48 } 49 while (i <= M) B[k++] = A[i++]; 50 while (j <= R) { 51 ans[A[j].id] += A[j].w * bit.query(A[j].y); 52 B[k++] = A[j++]; 53 } 54 for (int i=L; i<=R; ++i) { 55 bit.clear(A[i].y); 56 A[i] = B[i]; 57 } 58 } 59 int main() { 60 61 W = read();W = read(); // 把W在这里int了一下，然后。。。 62 int Index = 0,QueIndex = 0; 63 64 while (1) { 65 int opt = read(); 66 if (opt==3) break; 67 if (opt==1) { 68 int a = read(),b = read(),val = read(); 69 A[++Index] = (Que){0,a,b,val,0}; 70 } 71 else { 72 int a1 = read(),b1 = read(),a2 = read(),b2 = read(); 73 A[++Index] = (Que){1,a1-1,b1-1,1,++QueIndex}; 74 A[++Index] = (Que){1,a1-1,b2,-1,QueIndex}; 75 A[++Index] = (Que){1,a2,b1-1,-1,QueIndex}; 76 A[++Index] = (Que){1,a2,b2,1,QueIndex}; 77 } 78 } 79 CDQ(1,Index); 80 for (int i=1; i<=QueIndex; ++i) printf("%d ",ans[i]); 81 return 0; 82 }