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题意：一个n*m的矩阵，要么是 . 要么是 z ,问可以形成几个大z
分析：（直接奉上官方题解，我感觉说的实在是太好了）
Let‘s precalculate the values zlij, zrij, zldij — the maximal number of letters ‘z‘ to the left, to the right and to the left-down from the position (i, j). It‘s easy to do in O(nm) time. Let‘s fix some cell (i, j). Consider the value c = min(zlij, zldij). It‘s the maximum size of the square with upper right ceil in (i, j). But the number of z-patterns can be less than c. Consider some cell (x, y) diagonally down-left from (i, j) on the distance no more than c. The cells (i, j) and (x, y) forms z-pattern if y + zrxy > j.
Let‘s maintain some data structure for each antidiagonal (it can be described by formula x + y) that can increment in a point and take the sum on a segment (Fenwick tree will be the best choice for that). Let‘s iterate over columns j from the right to the left and process the events: we have some cell (x, y) for which y + zrxy - 1 = j. In that case we should increment the position y in the tree number x + y by one. Now we should iterate over the cells (x, y) in the current column and add to the answer the value of the sum on the segment from j - c + 1 to j in the tree number i + j .
补充:这样从右到左更新，当更新第j列的时候，每一条对角线上的 能够向右延伸到大于等于j的"z"都已经更新完毕，直接统计就行
时间复杂度O(nmlogm)
代码：
#include <cstdio> #include <iostream> #include <vector> using namespace std; typedef long long LL; const int N = 3e3+5; char s[N][N]; short l[N][N],r[N][N],x[N][N]; int n,m; short c[N*2][N]; void add(int x,int i) { for(; i<=m; i+=(i&(-i))) c[x][i]++; } int sum(int x,int i) { int res=0; for(; i>0; i-=(i&(-i))) res+=c[x][i]; return res; } struct Point { int x,y; }; vector<Point>g[N]; int main() { LL ans=0; scanf("%d%d",&n,&m); for(int i=1; i<=n; ++i) scanf("%s",s[i]+1); for(int i=1; i<=n; ++i) { for(int j=1,k=m; j<=m; ++j,--k) { if(s[i][j]==‘z‘)l[i][j]=l[i][j-1]+1; if(s[i][k]==‘z‘)r[i][k]=r[i][k+1]+1; } } for(int j=1; j<=m; ++j) { for(int i=1; i<=n; ++i) { if(s[i][j]==‘.‘)continue; x[i][j]=x[i+1][j-1]+1; } } for(int i=1; i<=n; ++i) for(int j=1; j<=m; ++j) if(s[i][j]==‘z‘) g[j+r[i][j]-1].push_back(Point {i,j}); for(int j=m; j>=1; --j) { for(int i=0; i<g[j].size(); ++i) { int t=g[j][i].x+g[j][i].y; add(t,g[j][i].y); } for(int i=1; i<=n; ++i) { if(s[i][j]==‘.‘)continue; int p=min(l[i][j],x[i][j]); ans+=sum(i+j,j)-sum(i+j,j-p); } } printf("%I64d\n",ans); return 0; }
Codeforces 628E Zbazi in Zeydabad 树状数组
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原文地址：http://www.cnblogs.com/shuguangzw/p/5215113.html
