Three Pieces CodeForces - 1065D (BFS)

链接

大意: n*n棋盘, 每个格子写有数字, 各不相同, 范围[1,n*n], 初始在数字1的位置, 可以操纵knight,bishop,rook三种棋子, 每走一步花费1, 交换棋子花费1, 问按顺序遍历完[1,n*n]的最少花费和最少花费下的最少交换次数

这题码了1个小时, WA on test 86, 没看出来是哪里错了, 不想改了

设状态(z,x,y)表示到达点(x,y)时棋子为z的最少花费, 每次BFS求出最短路, 再DP两次求出最少花费

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define x first
#define y second
using namespace std;
typedef pair<int,int> pii;

const int N = 15, INF = 0x3f3f3f3f;
int n;
pii a[N*N];
int f[N*N][3], g[N*N][3], d[N*N][3][3];
const int dx[]={-2,-1,1,2,2,1,-1,-2};
const int dy[]={1,2,2,1,-1,-2,-2,-1};
const int dx1[]={1,-1,1,-1};
const int dy1[]={1,1,-1,-1};
struct _ {int z,x,y;};
int dis[3][N][N], vis[3][N][N];

int chk(int x, int y) {
	return 1<=x&&x<=n&&1<=y&&y<=n;
}

int bfs(int z1, int x1, int y1, int z2, int x2, int y2) {
	queue<_> q;
	q.push({z1,x1,y1});
	memset(vis,0,sizeof vis);
	memset(dis,INF,sizeof dis);
	dis[z1][x1][y1] = 0;
	while (!q.empty()) {
		auto u = q.front();q.pop();
		if (vis[u.z][u.x][u.y]) continue;
		vis[u.z][u.x][u.y] = 1;
		int w = dis[u.z][u.x][u.y]+1;
		REP(i,0,7) {
			int xx=u.x+dx[i], yy=u.y+dy[i];
			if (!chk(xx,yy)) continue;
			dis[0][xx][yy] = min(dis[0][xx][yy], w+(u.z!=0));
			q.push({0,xx,yy});
		}
		REP(i,0,3) REP(j,1,n) {
			int xx=u.x+j*dx1[i],yy=u.y+j*dy1[i];
			if (!chk(xx,yy)) continue;
			dis[1][xx][yy] = min(dis[1][xx][yy], w+(u.z!=1));
			q.push({1,xx,yy});
		}
		REP(i,1,n) {
			dis[2][i][u.y] = min(dis[2][i][u.y], w+(u.z!=2));
			q.push({2,i,u.y});
			dis[2][u.x][i] = min(dis[2][u.x][i], w+(u.z!=2));
			q.push({2,u.x,i});
		}
	}
	return dis[z2][x2][y2];
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) REP(j,1,n) {
		int t;
		scanf("%d", &t);
		a[t] = {i,j};
	}
	REP(i,2,n*n) {
		REP(j,0,2) f[i][j] = INF;
		REP(j,0,2) REP(k,0,2) {
			d[i][j][k] = bfs(k,a[i-1].x,a[i-1].y,j,a[i].x,a[i].y);
			f[i][j] = min(f[i][j], f[i-1][k]+d[i][j][k]);
		}
	}
	REP(i,2,n*n) {
		REP(j,0,2) g[i][j] = INF;
		REP(j,0,2) REP(k,0,2) {
			if (f[i-1][k]+d[i][j][k]==f[i][j]) {
				g[i][j] = min(g[i][j], g[i-1][k]+(j!=k));
			}
		}
	}
	int ans1=INF, ans2=INF;
	REP(i,0,2) ans1=min(ans1,f[n*n][i]);
	REP(i,0,2) if (ans1==f[n*n][i]) ans2=min(ans2,g[n*n][i]);
	printf("%d %d\n", ans1, ans2);
}