codeforces 1065D

Posted 2021-01-16 buerdepepeqi

 题目链接：https://codeforces.com/problemset/problem/1065/D

题意：给你一个又1~n^2组成的n行n列的矩阵，你可以走日字型，直线，斜线，现在要求你从1走到n^2的位置，必须经过1，2，3，4and so on，请问你最短的路径长度和需要变换的方式

题解：非常明显的一个搜索，看了题解后我才明白要用dp存状态。。。。。

　　　这题可以说是一个比较好的把搜索和dp结合起来的题了，首先我们设置dp的状态为，当棋子处于坐标（i，j）时他的当前角色为z，他换了t种角色，并且已经走过前k个点

　　　然后就bfs搜索各种状态即可，首先把三种情况的状态存一下，然后，搜索当前走日字形的情况，搜索当前走直线的情况，搜索当前走斜线的情况，最后遍历所有状态中，走到终点时所有的情况即可

 

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
LL gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
    return a / gcd(a, b) * b;
}
LL powmod(LL a, LL b, LL MOD) {
    LL ans = 1;
    while(b) {
        if(b % 2)ans = ans * a % MOD;
        a = a * a % MOD;
        b /= 2;
    }
    return ans;
}
double dpow(double a, LL b) {
    double ans = 1.0;
    while(b) {
        if(b % 2)ans = ans * a;
        a = a * a;
        b /= 2;
    }
    return ans;
}

struct node {
    int x, y, z, t, k;
    //x，y 为坐标
    //z为当前方式
    //t为换了多少次
    //k为经过了前k个点
    node(int _x, int _y, int _z, int _t, int _k) {
        x = _x, y = _y, z = _z, t = _t, k = _k;
    }
};
int n;
int g[15][15];
int dp[15][15][3][205][205];
////0表示马，1表示车，2表示斜线
//int dx1[8][2]={{-2,-1},{-2,1},{2,1},{2,-1},{-1,-2},{-1,2},{1,-2},{1,2}};
//int dx2[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
//int dx3[4][2]={{-1,-1},{-1,1},{1,1},{1,-1}};

int dx1[8][2] = {{-2, -1}, {-2, 1}, {2, -1}, {2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}}; //日
int dx2[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}}; //直线
int dx3[4][2] = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}}; //斜线
int sx, sy;
int ex, ey;
void bfs(int sx, int sy) {
    memset(dp, -1, sizeof(dp));
    dp[sx][sy][0][0][1] = dp[sx][sy][1][0][1] = dp[sx][sy][2][0][1] = 0;
    queue<node>q;
    q.push(node(sx, sy, 0, 0, 1));
    q.push(node(sx, sy, 1, 0, 1));
    q.push(node(sx, sy, 2, 0, 1));
    while(!q.empty()) {
        node nd = q.front();
        q.pop();
        int x = nd.x, y = nd.y, z = nd.z, t = nd.t, k = nd.k;

        for(int i = 0; i < 3; i++) {
            if(i == z) continue;
            if(dp[x][y][i][t + 1][k] != -1) continue;
            dp[x][y][i][t + 1][k] = dp[x][y][z][t][k] + 1;
            q.push(node(x, y, i, t + 1, k));
        }
        //        cout<<x<<endl;
        if(z == 0) {
            for(int i = 0; i < 8; i++) {
                int nx = x + dx1[i][0];
                int ny = y + dx1[i][1];
                int nk = k;
                if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
                if(g[nx][ny] == k + 1) nk++;
                if(dp[nx][ny][z][t][nk] != -1) continue;
                dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
                q.push(node(nx, ny, z, t, nk));
            }
        }

        if(z == 1) {
            for(int j = 1; j <= 10; j++) {
                for(int i = 0; i < 4; i++) {
                    int nx = x + j * dx2[i][0];
                    int ny = y + j * dx2[i][1];
                    int nk = k;
                    if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
                    if(g[nx][ny] == k + 1) nk++;
                    if(dp[nx][ny][z][t][nk] != -1) continue;
                    dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
                    q.push(node(nx, ny, z, t, nk));
                }
            }
        }
        if(z == 2) {
            for(int j = 1; j <= 10; j++) {
                for(int i = 0; i < 4; i++) {
                    int nx = x + j * dx3[i][0];
                    int ny = y + j * dx3[i][1];
                    int nk = k;
                    if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
                    if(g[nx][ny] == k + 1) nk++;
                    if(dp[nx][ny][z][t][nk] != -1) continue;
                    dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
                    //q.push(node(nx, ny, z,, y, nk));
                    q.push(node(nx, ny, z, t, nk));
                }
            }
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    cin >> n;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            cin >> g[i][j];
            if(g[i][j] == 1) {
                sx = i;
                sy = j;
            }
            if(g[i][j] == n * n) {
                ex = i;
                ey = j;
            }
        }
    }
    bfs(sx, sy);
    int ans = INF;
    for(int z = 0; z < 3; z++) {
        for(int t = 0; t < 205; t++) {
            if(dp[ex][ey][z][t][n * n] != -1) {
                ans = min(dp[ex][ey][z][t][n * n], ans);
            }
        }
    }
    int flag = 0;
    for(int t = 0; t < 205; t++) {
        for(int z = 0; z < 3; z++) {
            if(dp[ex][ey][z][t][n * n] == ans && !flag) {
                cout << ans << " " << t << endl;
                flag = 1;
            }
        }
    }
}

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