bzoj3939 [Usaco2015 Feb]Cow Hopscotch

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3939

【题解】

f[i][j] = s[i-1][j-1] - sum[a[i][j]]

用cdq分治来处理横坐标，先处理上面的dp值，再讨论上面部分对下面的贡献，然后分治处理下面的dp值即可。

如果加了快读好像排bzoj rk1？

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 7e5 + 10, N = 750 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, m, K;
int a[N][N], t[M], f[N][N];

inline void solve(int l, int r) {
    if(l == r) return ;
    int mid = l+r>>1, sum = 0;
    solve(l, mid);
    for (int j=1; j<=m; ++j) {
        for (int i=mid+1; i<=r; ++i) {
            f[i][j] += (sum - t[a[i][j]] + mod) % mod;
            if(f[i][j] >= mod) f[i][j] -= mod;
        }
        for (int i=l; i<=mid; ++i) {
            sum = sum + f[i][j];
            if(sum >= mod) sum -= mod;
            t[a[i][j]] = t[a[i][j]] + f[i][j];
            if(t[a[i][j]] >= mod) t[a[i][j]] -= mod;
        }
    }
    for (int j=1; j<=m; ++j)
        for (int i=l; i<=mid; ++i)
            t[a[i][j]] = 0;
    solve(mid+1, r);
}

int main() {
    cin >> n >> m >> K;
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=m; ++j) scanf("%d", &a[i][j]);
    f[1][1] = 1;
    solve(1, n);
    cout << f[n][m];
    return 0;
}