poj-3083-烦人复杂搜索+bfs,dfs模板

来源：互联网发布：mac app store 很慢编辑：程序博客网时间：2024/05/21 07:52

仅仅记录，本题无意义仅仅是烦人，当以后模板吧
ps：其实可以不写这么多的从终点沿墙往右走跟从起点沿墙往左走是一个意思的。。。AC后想通的。。。哎思路还是太窄。
求条件是否成立用dfs较好，求最优解bfs最好。

#include<cstdio>#include<queue>using namespace std;char mp[50][50];struct node{    int x,y,dis;}no1,no2,star,en;int bu[4][2] = {0,1,0,-1,1,0,-1,0};int viewleft[4][2] = {-1,0,0,1,1,0,0,-1};int left[4][2] =     {0,-1,-1,0,0,1,1,0};int viewright[4][2] = {-1,0,0,1,1,0,0,-1};int right[4][2] =     {0,1,1,0,0,-1,-1,0};int m,n;int dfsl(node pre,int view){    if(pre.x==en.x&&pre.y==en.y)return pre.dis;    int x = pre.x+left[view][0],y = pre.y+left[view][1];    if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]!='#'){        no1.x = x;        no1.y = y;        no1.dis = pre.dis+1;        return dfsl(no1,(view-1+4)%4);      }    else {        int x = pre.x+viewleft[view][0],y = pre.y+viewleft[view][1];        if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]!='#'){        no1.x = x;        no1.y = y;        no1.dis = pre.dis+1;        return dfsl(no1,view);        }        else if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]=='#')            return dfsl(pre,(view+1)%4);    }}//沿着左走 int dfsr(node pre,int view){  if(pre.x==en.x&&pre.y==en.y)return pre.dis;    int x = pre.x+right[view][0],y = pre.y+right[view][1];    if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]!='#'){        no1.x = x;        no1.y = y;        no1.dis = pre.dis+1;        return dfsr(no1,(view+1)%4);      }    else {        int x = pre.x+viewright[view][0],y = pre.y+viewright[view][1];        if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]!='#'){        no1.x = x;        no1.y = y;        no1.dis = pre.dis+1;        return dfsr(no1,view);        }        else if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]=='#')            return dfsr(pre,(view-1+4)%4);    }}//沿着右边 int bfs(){    int co[50][50] ={0};    queue<node>qu;    star.dis = 0;    qu.push(star);    co[star.x][star.y] = 1;    while(!qu.empty())    {        no1 = qu.front();        qu.pop();        if(no1.x==en.x&&no1.y==en.y)return no1.dis;        for(int i=0;i<4;i++)        {            no2.x = no1.x+bu[i][0];            no2.y = no1.y+bu[i][1];            if(no2.x>=0&&no2.x<n&&no2.y>=0&&no2.y<m&&!co[no2.x][no2.y]&&mp[no2.x][no2.y]!='#')            {                no2.dis = no1.dis+1;                qu.push(no2);                co[no2.x][no2.y] = 1;            }        }    }}//求最短路径。。。 int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&m,&n);        for(int i=0;i<n;i++)          scanf("%s",mp[i]);        for(int i=0;i<n;i++)         for(int j=0;j<m;j++)          {            if(mp[i][j]=='S'){star.x = i;star.y = j;}            if(mp[i][j]=='E'){en.x = i;en.y = j;}          }        int view;        if(star.x==0)view = 2;        else if(star.x==n-1)view = 0;        else if(star.y==0)view = 1;        else view = 3;        printf("%d ",dfsl(star,view)+1);        printf("%d ",dfsr(star,view)+1);        printf("%d\n",bfs()+1);    }    return 0;}

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