走迷宫

来源：互联网发布：文件md5加密 java 编辑：程序博客网时间：2024/05/01 18:52

1.迷宫中是否存在一条路从(x1,y1)到 (x2,y2),矩阵为1或0

#include <iostream> #include <cstdio>#include <cstring>using namespace std;int t,n,m,x1,y1,x2,y2;int in[1000][1000];int mx[4]={-1,1,0,0};int my[4]={0,0,-1,1};int dfs(int x,int y) //深搜{    if(x==x2&&y==y2)      return 1;    for(int i=0;i<4;i++)    {        int x3 = x+mx[i];        int y3 = y+my[i];        if(x3&&x3<=n &&y3&&y3<=m &&in[x3][y3]){            in[x3][y3] = 0;            if(dfs(x3,y3)) return 1;        }  //走过(x3,y3)时,令in[x3][y3]=0表示已走过该点下次不会再重复走，如果dfs(x3,y3)=1,表示通过这个点能走到终点    }    return 0;}int main(){    cin >> t;    while(t--)     {        cin>>n>>m;        for(int i=1;i<=n;i++)         for(int j=1;j<=m;j++)           cin>>in[i][j];        cin>>x1>>y1>>x2>>y2;        in[x1][y1]=0;        cout<<dfs(x1,y1)<<endl;    }    return 0;}

2.能否走出迷宫以及最短路（左上角和右下角为1，即入口和出口)

#include<iostream>#include<queue>  //广搜using namespace std;int maze[100][100];int move[4][2]={{1,0},{-1,0},{0,1},{0,-1}};struct point {     int x, y;         int step; };int main() {    int n;    point tem, next;    queue<point> q;    while(cin>>n,n) {        while(!q.empty()) q.pop();        for(int i=0; i<n; i++)           for(int j=0; j<n; j++)             cin >> maze[i][j];        maze[0][0] = 0;                       tem.x = tem.y = 0;        tem.step = 1;        q.push(tem); //起点入队        while(!q.empty()) {                         tem = q.front();//取队头            if (tem.x==n-1 && tem.y==n-1) break; //出口for(int i=0;i<4;i++){  //将队头下步可走的点全部入队int x = tem.x+move[i][0];int y = tem.y+move[i][1];    if(x>=0&&x<n&&y>=0&&y<n &&maze[x][y]){   maze[x][y] = 0;next.x = x;    next.y = y;    next.step = tem.step + 1;    q.push(next);    }}             q.pop();//将队头出队，继续下一步判断         }        if (tem.x == n-1 && tem.y == n-1)            cout << q.front().step << endl;        else cout << 0 << endl;                  }    return 0;}

3.走出迷宫最小权重（所有路都可以走）

#include<iostream>#include<cstdio>#include <cstring>#include <queue>#define INF 10001using namespace std;int n,m;int g[101][101];bool v[101][101]; int move[4][2]={{0,1},{0,-1},{1,0}, {-1,0}};struct pot{    int x,y,sum;    pot(int a,int b,int c):x(a),y(b),sum(c){}    friend bool operator < (pot a,pot b){ //自定义优先级        return a.sum > b.sum;  // sum最小的在top()里    } };int dij(pot a,pot b){    priority_queue<pot> q;//优先级队列    memset(v,false,sizeof(v));    v[a.x][a.y]=true;pot tmp = a;    while(!v[b.x][b.y])    {        for(int i=0;i<4;i++){            int x = tmp.x + move[i][0];            int y = tmp.y + move[i][1];            if(x&&x<=n &&y&&y<=m &&!v[x][y]){            pot next(x,y,tmp.sum+g[x][y]);            q.push(next);            }        }        tmp = q.top();        while(v[tmp.x][tmp.y]){//更新最短路径            q.pop();            tmp = q.top();        }        v[tmp.x][tmp.y] = true;      }    return tmp.sum ;}int main(){    int t,x1,y1,x2,y2;    scanf("%d",&t);    while(t--)    {                                                                      scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)          for(int j=1;j<=m;j++)             scanf("%d",&g[i][j]);        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);        pot a(x1,y1,g[x1][y1]);        pot b(x2,y2,g[x2][y2]);        cout << dij(a,b) << endl;   }    return 0;}

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