題意

題目連結

給出一個\(n \times n\)的矩陣，允許修改\(k\)次，求一條從\((1, 1)\)到\((n, n)\)的路徑。要求字典序最小

Sol

很顯然的一個思路是對於每個點，預處理出從\((1, 1)\)到該點最多能經過多少個\(1\)

然後到這裡我就不會做了。。

接下來應該還是比較套路的吧，就是類似於BFS一樣，可以列舉步數，然後再列舉向下走了幾次，有點分層圖的感覺。

/*
*/
#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define LL long long 
#define rg register 
#define pt(x) printf("%d ", x);
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2001, INF = 1e9 + 10, mod = 1e9 + 7, B = 1;
const double eps = 1e-9;
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 N, K, vis[MAXN][MAXN], f[MAXN][MAXN];
char s[MAXN][MAXN];
int main() {
    N = read(); K = read();
    for(int i = 1; i <= N; i++) scanf("%s", s[i] + 1);
    for(int i = 1; i <= N; i++) {
        for(int j = 1; j <= N; j++) {
            f[i][j] = max(f[i - 1][j], f[i][j - 1]);
            if(s[i][j] == 'a') f[i][j]++;
            if(i + j - 1 - f[i][j] <= K) s[i][j] = 'a';
        }
    }
    putchar(s[1][1]); vis[1][1] = 1;
    for(int i = 3; i <= N << 1; i++) {//all step
        char now = 'z' + 1;
        for(int j = 1; j < i; j++) { // num of down
            if(j > N || (i - j > N)) continue;
            if(!vis[j - 1][i - j] && !vis[j][i - j - 1]) continue;
            now = min(now, s[j][i - j]);    
        }
        putchar(now);
        for(int j = 1; j < i; j++) { // num of down
            if(j > N || (i - j > N)) continue;
            if(!vis[j - 1][i - j] && !vis[j][i - j - 1]) continue;
            if(s[j][i - j] == now) vis[j][i - j] = 1;   
        }       
    }
    return 0;
}
/*

*/