题目描述
White Rabbit has a rectangular farmland of n*m. In each of the grid there is a kind of plant. The plant in the j-th column of the i-th row belongs the a[i][j]-th type.
White Cloud wants to help White Rabbit fertilize plants, but the i-th plant can only adapt to the i-th fertilizer. If the j-th fertilizer is applied to the i-th plant (i!=j), the plant will immediately die.
Now White Cloud plans to apply fertilizers T times. In the i-th plan, White Cloud will use k[i]-th fertilizer to fertilize all the plants in a rectangle [x1[i]...x2[i]][y1[i]...y2[i]].
White rabbits wants to know how many plants would eventually die if they were to be fertilized according to the expected schedule of White Cloud.
White Cloud wants to help White Rabbit fertilize plants, but the i-th plant can only adapt to the i-th fertilizer. If the j-th fertilizer is applied to the i-th plant (i!=j), the plant will immediately die.
Now White Cloud plans to apply fertilizers T times. In the i-th plan, White Cloud will use k[i]-th fertilizer to fertilize all the plants in a rectangle [x1[i]...x2[i]][y1[i]...y2[i]].
White rabbits wants to know how many plants would eventually die if they were to be fertilized according to the expected schedule of White Cloud.
输入描述:
The first line of input contains 3 integers n,m,T(n*m<=1000000,T<=1000000)
For the next n lines, each line contains m integers in range[1,n*m] denoting the type of plant in each grid.
For the next T lines, the i-th line contains 5 integers x1,y1,x2,y2,k(1<=x1<=x2<=n,1<=y1<=y2<=m,1<=k<=n*m)
输出描述:
Print an integer, denoting the number of plants which would die.
示例1
输出
复制3
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 typedef long long LL; 6 7 //c[i] = a[i]-a[i-1] 8 //c2[i] = (i-1)*c[i] 9 //树状数组维护c,c2 10 //ans[1-n]: n*sigma(c,n) - sigma(c2,n) 11 const int N = 100000010; 12 #define lowbit(i) (i&(-i)) 13 LL c[N],c2[N]; 14 LL n; 15 void up(LL *r,LL x,LL val){ 16 while(x<=n){ 17 r[x] += val; 18 x += lowbit(x); 19 } 20 } 21 LL down(LL *r,LL x){ 22 LL ans = 0; 23 while(x){ 24 ans += r[x]; 25 x -= lowbit(x); 26 } 27 return ans; 28 } 29 30 int main() 31 { 32 LL q; 33 LL a,b,f; 34 scanf("%lld",&n); 35 scanf("%lld",&q); 36 for(int i = 1; i <= n; i++){ 37 LL f; 38 scanf("%lld",&f); 39 up(c,i,f); up(c,i+1,-f); 40 up(c2,i,f*(i-1)); up(c2,i+1,-f*(i)); 41 } 42 43 while(q--){ 44 int type; 45 scanf("%d",&type); 46 47 if(type==1){ 48 scanf("%lld%lld%lld",&a,&b,&f); 49 up(c,a,-f); up(c,b+1,f); 50 up(c2,a,-f*(a-1)); up(c2,b+1,f*b); 51 52 } 53 else{ 54 scanf("%lld%lld%lld",&a,&b,&f); 55 up(c,a,f); up(c,b+1,-f); 56 up(c2,a,f*(a-1)); up(c2,b+1,-f*b); 57 } 58 59 } 60 scanf("%lld%lld",&a,&b); 61 LL l = (a-1)*down(c,a-1) - down(c2,a-1); 62 LL r = b*down(c,b) - down(c2,b); 63 printf("%lld ",r-l); 64 return 0; 65 }