题目

思路

本思路参考博客AcWing 471. 棋盘 - AcWing

约束方程： 

代码

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 110, INF = 0x3f3f3f3f;
int g[N][N], n, m, dist[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};void dfs(int x, int y, int cost, int used) 
{if (dist[x][y] <= cost) return ; //花费更少，说明没必要DFS，直接剪枝dist[x][y] = cost;if (x == m && y == m) return ;for (int i = 0; i < 4; i ++ ) //四个方向深搜{int nx = x + dx[i], ny = dy[i] + y;if (nx < 1 || nx > m || ny < 1 || ny > m) continue ;if (g[nx][ny] == -1)  //无颜色{if (used) continue ; //已经使用魔法的不能再次使用else {g[nx][ny] = g[x][y];dfs(nx, ny, cost + 2, 1);g[nx][ny] = -1;}}else if (g[nx][ny] == g[x][y]) dfs(nx, ny, cost, 0);//颜色相同else dfs(nx, ny, cost + 1, 0);//颜色不同，使用魔法}
}int main() 
{scanf("%d%d", &m, &n);memset(g, -1, sizeof g);while (n -- ) {int a, b, c;scanf("%d%d%d", &a, &b, &c);g[a][b] = c;}memset(dist, 0x3f, sizeof dist);dfs(1, 1, 0, 0);if (dist[m][m] == INF) puts("-1");else printf("%d\n", dist[m][m]);return 0;
}

xmuoj：xmuoj | 璃月森林探险：符文之路