走迷宫值路径记录

来源：互联网发布：网络代理合同范本编辑：程序博客网时间：2024/05/16 14:58

利用bfs走迷宫是最基础的方法，记录访问过的每个点然后回溯最短路径的点

首先把问题简单化一下，只有墙壁和路，代码比较简单，只要会bfs基本上看得懂

// 走迷宫记录路径// 概述：有一个N*M大小的迷宫，其中'#'代表墙壁无法通过，'.'代表路，每走一步都将花费1s的时间，求从起点(0,0)到终点(N-1,M-1)的最短路径，并将路径打印#include <iostream>#include <queue>#include <cstring>using namespace std;const int INF = 1000005;const int MAX_N = 105;struct Point{    int x;    int y;}first, next,pre[MAX_N][MAX_N];char pic[MAX_N][MAX_N];int N, M;int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};int value[MAX_N][MAX_N];bool vis[MAX_N][MAX_N];int bfs(){    queue<Point> que;    que.push(first);    while(que.size())    {        first = que.front();        que.pop();        if(first.x == N - 1 && first.y == M -1)            return value[N - 1][M - 1];        for(int i = 0; i < 4; i++)        {            next.x = first.x + dx[i];            next.y = first.y + dy[i];            if(next.x >= 0 && next.x < N && next.y >= 0 && next.y < M && pic[next.x][next.y] != '#' && vis[next.x][next.y])            {                value[next.x][next.y] = value[first.x][first.y] + 1;                pre[next.x][next.y].x = first.x;                pre[next.x][next.y].y = first.y;                vis[next.x][next.y] = false;                que.push(next);            }        }    }    return INF;}void print_pre(int i, int j){    if(pre[i][j].x == -1)        return;    print_pre(pre[i][j].x, pre[i][j].y);    printf("%ds:(%d,%d)->(%d,%d)\n",value[i][j], pre[i][j].x, pre[i][j].y, i, j);}int main(){    while(cin >> N >> M)    {        for(int i = 0; i < N; i++)            cin >> pic[i];        first.x = 0;        first.y = 0;        memset(vis, true, sizeof(vis));     // true表示该点未访问，false表示该点已访问        vis[0][0] = false;        pre[0][0].x = pre[0][0].y = -1;        int ans = bfs();        if(ans == INF)            printf("sorry,can't out!\n");        else        {            printf("It takes %d seconds to reach the target position, let me show you the way.\n", ans);            print_pre(N - 1, M - 1);        }    }    return 0;}

hangd1026题就是这种类型的题目，只是多了一个条件，路上可能会有怪物，每个怪物有N点生命值（1<=n<=9），要花费n秒杀死这个怪物。

题目链接

#include <iostream>#include <queue>#include <cstring>using namespace std;const int INF = 100005;const int MAX_N = 105;struct Node{    int x;    int y;    int num;    bool operator<(const Node &a) const    {        return num > a.num;    }}first, nextNode;struct Point{    int x;    int y;}pre[MAX_N][MAX_N];char pic[MAX_N][MAX_N];int value[MAX_N][MAX_N];bool vis[MAX_N][MAX_N];int N, M;int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};int bfs(){    priority_queue<Node> pque;    pque.push(first);    while(pque.size())    {        first = pque.top();        pque.pop();        if(first.x == N - 1 && first.y == M - 1)            return value[N - 1][M - 1];        for(int i = 0; i < 4; i++)        {            nextNode.x = first.x + dx[i];            nextNode.y = first.y + dy[i];            if(nextNode.x >= 0 && nextNode.x < N && nextNode.y >= 0 && nextNode.y < M && pic[nextNode.x][nextNode.y] != 'X' && vis[nextNode.x][nextNode.y])            {                if(pic[nextNode.x][nextNode.y] == '.')                {                    nextNode.num = first.num + 1;                }                else if(isdigit(pic[nextNode.x][nextNode.y]))                {                    nextNode.num = first.num + 1 + (int)(pic[nextNode.x][nextNode.y] - '0');                }                pre[nextNode.x][nextNode.y].x = first.x;                pre[nextNode.x][nextNode.y].y = first.y;                value[nextNode.x][nextNode.y] = nextNode.num;                vis[nextNode.x][nextNode.y] = false;                pque.push(nextNode);            }        }    }    return INF;}void print_pre(int i, int j){    if(pre[i][j].x == -1)        return;    print_pre(pre[i][j].x, pre[i][j].y);    if(pic[i][j] == '.')        printf("%ds:(%d,%d)->(%d,%d)\n",value[i][j], pre[i][j].x, pre[i][j].y, i, j);    else    {        int t = pic[i][j] - '0';        int kase = value[pre[i][j].x][pre[i][j].y];        printf("%ds:(%d,%d)->(%d,%d)\n",++kase, pre[i][j].x, pre[i][j].y, i, j);        for(int k = 0; k < t; k++)        {            printf("%ds:FIGHT AT (%d,%d)\n", ++kase, i, j);        }    }}int main(){    while(cin >> N >> M)    {        for(int i = 0; i < N; i++)            cin >> pic[i];        memset(vis, true, sizeof(vis));        first.x = 0;        first.y = 0;        first.num = 0;        value[0][0] = 0;        vis[0][0] = false;        pre[0][0].x = pre[0][0].y = -1;        int ans = bfs();        if(ans != INF)        {            printf("It takes %d seconds to reach the target position, let me show you the way.\n", ans);            print_pre(N - 1, M - 1);        }        else            printf("God please help our poor hero.\n");        printf("FINISH\n");    }    return 0;}

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