闲着无聊刷道题

Description

You are given a chessboard made up of N squares by M squares. Some of the squares are colored black, and the others are colored white. Please 
write a program to calculate the number of rectangles which are completely made up of white squares.

Input

There are multiple test cases. Each test case begins with two integer N,M (1 <= N , M<= 2000), the board size. The following N lines, each with M
characters, have only two valid character values:
b - representing a black square;
w - representing a white square.
Process to the end of file.

Output

For each test case in the input, output the number of white rectangles a line.

Sample Input

2 3
bbb
www
2 2
bw
wb

Sample Output

6
2

Hint

Source

The 4th Wuhan University Programming Contest(Preliminary Round) 

坑爹的是把数组赋初值改成memset就tle了，我去，我说栈优化O（n）复杂度的算法怎么都不可能tle啊！这oj真没良心。。

#include<iostream>
#include<string>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
#define pb push_back
#define MM(name,what) memset(name,what,sizeof(name))
typedef long long i64;
const int maxn = 2010;
const int inf = 0x3f3f3f3f;

struct zz
{
    int x;
    int re;
}zx;

int h[maxn][maxn];
char c[maxn][maxn];
int n,m;
vector<zz>s;

i64 gao(int id)
{
    i64 ans = 0;
    int temp = 0;
    s.clear();
    zx.x = -1;
    zx.re = 0;
    s.pb(zx);
    for(int i=1;i<=n;i++)
    {
        zx.x = h[i][id];
        zx.re = 1;
        while(zx.x < s.back().x)
        {
            temp -= s.back().x*s.back().re;
            zx.re += s.back().re;
            s.pop_back();
        }
        temp+=zx.x*zx.re;
        s.pb(zx);
        ans+=temp;
    }
    return ans;
}

i64 start()
{
    i64 ans=0;
    for(int i=0;i<=n;i++)
        for(int j=0;j<=m;j++)
            h[i][j]=0;
//  MM(h,0);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(c[i][j]=='w') h[i][j]=h[i][j-1]+1;

    for(int j=1;j<=m;j++)
    {
        ans+=gao(j);
    }
    return ans;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        getchar();
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                c[i][j] = getchar();
            }
            getchar();
        }
       // cout<<start()<<endl;
        printf("%I64d\n",start());
    }
    return 0;
}