搜索与图论2

宽度优先搜索BFS

queue<int> q;
st[1] = true; // 表示1号点已经被遍历过
q.push(1);

while (q.size())
{
    int t = q.front();
    q.pop();

    for (int i = h[t]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (!st[j])
        {
            st[j] = true; // 表示点j已经被遍历过
            q.push(j);
        }
    }
}

#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> PII;

const int N = 1e2 + 7;
int g[N][N], d[N][N];
int n, m;


int bfs() {
    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
    queue <PII> q;

    for (auto &v : d) 
        for (auto &x : v) {
            x = - 1;
        }

    d[0][0] = 0;
    q.push({0, 0});

    while (!q.empty()) {
        auto t = q.front();
        q.pop();

        for (int i = 0; i < 4; i++) {
            int x = t.first + dx[i], y = t.second + dy[i];

            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) {
                d[x][y] = d[t.first][t.second]  + 1;
                q.push({x, y});
            }
        }

    }


        return d[n - 1][m - 1];
}


int main() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> g[i][j];
        }
    }

    cout << bfs() << endl;

    return 0;
}

#include<stdio.h>
#include<queue>
#include<iostream>
using namespace std;
char maze[201][201];
int sx, sy, tx, ty;
//左右上下4个方向
int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };
int m, n;

struct node {
    int x,  y, dis;
};

bool operator < (const node & a, const node & b) {
    return a.dis > b.dis;
}

void bfs() {
    priority_queue<node> que;
    node st { sx,sy,0 };
    maze[sx][sy] = '#';
    que.push(st);

    while (!que.empty()) {
        node p = que.top();
        que.pop();
        //若已找到，则退出
        if (p.x == tx && p.y == ty) {
            cout << p.dis << endl;
            return;
        }
        for (int i = 0; i < 4; ++i) {
            int nx = p.x + dx[i];
            int ny = p.y + dy[i];
            node np{ nx,ny, 0};

            if (nx >= 0 && nx < n&&ny >= 0 && ny < m&&maze[nx][ny] != '#') {
                if (maze[nx][ny] == 'X')
                    np.dis = p.dis + 2;
                else
                    np.dis = p.dis + 1;
                maze[np.x][np.y] = '#';
                que.push(np);

            }
        }
    }
    printf("impossible\n");
}
int main() {
    while (cin>>n>>m) {
        for (int i = 0; i < n; i++)
            scanf("%s", maze[i]);
        for(int i=0; i<n; i++)
            for (int j = 0; j < m; j++) {
                if (maze[i][j] == 'S')
                    sx = i, sy = j;
                else if (maze[i][j] == 'T')
                    tx = i, ty = j;
            }
        bfs();
    }
    return 0;
}

八数码

#include<cstring>
#include<iostream>
#include<queue>
using namespace std;

int fact[9];
bool vis[362880];

int permutation_hash(char s[])          //求长度为9的字符串某种排列的哈希值
{
    int ans = 0;
    for(int i = 0; i < 9; i ++)
    {
        int d = 0;
        for(int j = 0; j < i; j ++)
            if(s[j] > s[i])  d ++;      //求s[i]与其前面的字符组成的逆序对个数
        ans += d * fact[i];
    }
    return ans;
}

typedef struct{
    char s[10];
    int step;
    int k;          //'x'在第k位
}Point;

int dx[4] = {-1, 0, 1, 0};
int dy[4] = { 0,-1, 0, 1};
int bfs(Point p)
{
    vis[permutation_hash(p.s)] = true;
    queue<Point> q;
    q.push(p);
    while(!q.empty())
    {
        p = q.front();
        q.pop();
        /*
        printf("%d ",p.step);       //print调试法
        puts(p.s);
        */
        if(!strcmp(p.s , "12345678x"))  return p.step;

        int x = p.k / 3;      //'x'的行数
        int y = p.k % 3;      //'x'的列数
        Point next;
        next.step = p.step + 1;
        for(int i = 0; i < 4; i ++)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if(nx >= 0 && nx <= 2 && ny >= 0 && ny <= 2)
            {
                next.k = nx * 3 + ny;           //求出'x'在字符串中的的新位置

                strcpy(next.s, p.s);
                next.s[9] = 0;
                next.s[p.k] = p.s[next.k];      //先用即将和'x'交换的字符覆盖'x'之前的位置
                next.s[next.k] = 'x';           //再给'x'的新位置赋值'x'

                int hash = permutation_hash(next.s);
                if(!vis[hash])
                {
                    vis[hash] = true;
                    q.push(next);
                }
            }
        }
    }
    return -1;
}

int main()
{
    fact[0] = 1;
    for(int i = 1; i < 9; i ++)  fact[i] = fact[i - 1] * i;    //预处理fact[i] = i!

    char c[2],str[10];
    Point start;
    for(int i = 0; i < 9; i ++)
    {
        scanf("%s",&c);
        if(c[0] == 'x')  start.k = i;
        start.s[i] = c[0];
    }
    start.s[9] = 0;
    start.step = 0;
    printf("%d",bfs(start));
    return 0;
}