题意

题目链接

Sol

\(f[i][j]\)表示前\(i\)行\(j\)列的贡献，转移的时候枚举从哪里转移而来，复杂度\(O(n^4)\)

然后考虑每一行的贡献，动态开节点线段树维护一下每种颜色的答案

转移的时候用总的方案减去相同颜色的方案

复杂度\(O(n^2 log^2 n)\)

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 1001, INF = 1e9 + 10, mod = 1000000007, SS = MAXN * MAXN * 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int add(int x, int y) {if(x + y < 0) return x + y + mod;return x + y >= mod ? x + y - mod : x + y;
}
int N, M, K, c[MAXN][MAXN], f[MAXN][MAXN], tot;
int root[SS], ls[SS], rs[SS], sum[SS];
void IntAdd(int &k, int l, int r, int pos, int val) {if(!k) k = ++tot;sum[k] = add(sum[k], val);if(l == r) return ;int mid = l + r >> 1;if(pos <= mid) IntAdd(ls[k], l, mid, pos, val);else IntAdd(rs[k], mid + 1, r, pos, val);
}
int Query(int k, int l, int r, int ll, int rr) {if(!k) return 0;if(ll <= l && r <= rr) return sum[k];int mid = l + r >> 1;if(ll > mid) return Query(rs[k], mid + 1, r, ll, rr);else if(rr <= mid) return Query(ls[k], l, mid, ll, rr);else return add(Query(ls[k], l, mid, ll, rr), Query(rs[k], mid + 1, r, ll, rr));
}
signed main() {N = read(); M = read(); K = read();f[1][1] = 1;for(int i = 1; i <= N; i++) {for(int j = 1; j <= M; j++) {int k = read(); c[i][j] = k;if(!f[i][j]) f[i][j] = add(Query(root[0], 1, M, 1, j - 1), -Query(root[k], 1, M, 1, j - 1));}for(int j = 1; j <= M; j++)IntAdd(root[0], 1, M, j, f[i][j]),IntAdd(root[c[i][j]], 1, M, j, f[i][j]);}cout << f[N][M];return 0;
}