hiho_99_骑士问题

Posted 2020-07-15 农民伯伯-Coding

题目大意

    给定国际象棋8x8棋盘上三个起始点，三个骑士分别从三个起始点开始移动（骑士只能走日字，且骑士从任意一点出发可以走遍整个棋盘）。现要求三个骑士汇聚到棋盘上某个点，且使得骑士到达该点所移动的次数总和最小。求该最小移动次数。 
题目连接:骑士问题

题目分析

    典型的搜索，最短路径可以使用BFS。骑士数只有三个，因此可以求出每个骑士到达棋盘上所有点的移动的次数，再遍历一遍棋盘，求出最小次数和。

实现

#pragma once
#pragma execution_character_set("utf-8")
// 本文件为utf-8 编码格式
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<unordered_map>
#include<unordered_set>
#include<string>
#include<stack>
#include<queue>
using namespace std;
struct Node{
	int x, y, step;
	Node(int xx, int yy, int s) :x(xx), y(yy), step(s){};
};
int move_step[9][9][3];
bool visited[9][9];
int move_inc[8][2] = { { -2, -1 }, { -2, 1 }, { -1, 2 }, { 1, 2 }, { 2, 1 }, { 2, -1 }, { 1, -2 }, { -1, -2 } };
void Bfs(int start_x, int start_y, int knight){
	memset(visited, false, sizeof(visited));
	queue<Node> Q;
	Node node(start_x, start_y, 0);
	Q.push(node);
	visited[start_x][start_y] = true;
	while (!Q.empty()){
		node = Q.front();
		Q.pop();
		move_step[node.x][node.y][knight] = node.step;
		for (int i = 0; i < 8; i++){
			int next_x = node.x + move_inc[i][0];
			int next_y = node.y + move_inc[i][1];
			if (next_x >= 1 && next_x <= 8 &&
				next_y >= 1 && next_y <= 8 &&
				!visited[next_x][next_y]){
				visited[next_x][next_y] = true;
				Q.push(Node(next_x, next_y, node.step + 1));
			}
		}
	}
}
void Init(){
	memset(move_step, -1, sizeof(move_step));
}
int MinStep(){
	int min = 1 << 30;
	for (int r = 1; r <= 8; r++){
		for (int c = 1; c <= 8; c++){
			min = min < (move_step[r][c][0] + move_step[r][c][1] + move_step[r][c][2]) ?
			min : (move_step[r][c][0] + move_step[r][c][1] + move_step[r][c][2]);
		}
	}
	return min;
}
int main(){
	int T;
	char row, col;
	scanf("%d", &T);
	int start_x[3];
	int start_y[3];
	
	while (T--){
		for (int i = 0; i < 3; i++){
			getchar();
			scanf("%c%c", &row, &col);			
			start_x[i] = row - ‘A‘ + 1;
			start_y[i] = col - ‘0‘;
		}
		for (int i = 0; i < 3; i++){
			Bfs(start_x[i], start_y[i], i);
		}
		int result = MinStep();
		printf("%d\n", result);
	}
	return 0;
}