Uva439：BFS题目总结

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <string>
using namespace std;

const int maxn = 500 + 20;
int map[9][9]; 
int cnt;
int d[9][9];
string start, over;
int dir[8][2] = { { -2, 1 }, { -1, 2 }, { 1, 2 }, { 2, 1 }, { 2, -1 }, { 1, -2 }, { -1, -2 }, { -2, -1 } };
　　//棋盘方向通常会用二维数组来表示方向
struct Node
{
    int r, c;    // 行,列
    //int dir;     // 方向
    Node(int r = 0, int c = 0) : r(r), c(c) {}
};

bool inside(int r, int c)
{
    if (r < 0 || r >= 8 || c < 0 || c >= 8) return false;
    return true;
}

//start row, start col, over row, over col
bool BFS(int sr, int sc, int or, int oc)
{
    queue<Node> Q;
    memset(d, 0, sizeof(d));
    Node u(sr, sc);
    Q.push(u);
    while (!Q.empty()) {
        Node u = Q.front(); Q.pop();
        if ((u.r == or) && (u.c == oc)) {      //如果找到，就true;
            return true;
        }
        for (int i = 0; i < 8; i++) {
            Node v = Node(u.r + dir[i][0], u.c + dir[i][1]);
            if (inside(v.r, v.c) && d[v.r][v.c] == 0) {
                d[v.r][v.c] = d[u.r][u.c] + 1;
                Q.push(v);
            }
        }
    }
    return false;
}

void print_ans(Node u)
{
    cout << "To get from " << start << " to " << over << " takes " << d[u.r][u.c] << " knight moves.
";
}

int main()
{
    while (cin >> start >> over) {
        memset(map, 0, sizeof(map));
        
        int sr = start[0] - 'a', sc = start[1] - '0' - 1, or = over[0] - 'a', oc = over[1] - '0' - 1;
        BFS(sr, sc, or, oc);
        print_ans(Node(or, oc));
    }
    return 0;
}

总结：通常最短路（最优解）问题会考虑用BFS来进行求解