POJ 3009 Curling 2.0（dfs）

来源：互联网发布：画图软件如何使用方法编辑：程序博客网时间：2024/06/09 19:47

题目意思就是一个冰壶，要从起点到达终点，你推他一把，他就沿着直线开始滑行，直到撞到一块石头，撞到石头冰壶停止，石头也会消失。但冰壶不会取代那个石头的位置，而是在原来碰撞的位置，问你10次之内能不能到达终点。

用dfs+回溯的方法暴力去搜，顺带剪枝，如果答案大于10了，就不用搜了，这个一定要剪，否则TLE。。。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define INT_MAX 1<<30#define debug cout<<xixi.time<<j<<endl;int w,h;int sx,sy;char maze[21][21];int ex,ey;bool flag;bool check;int ans;int dir;struct wode{    int x, y;    int time;};bool can_not(int x, int y){    if(x<0||x>=h||y>=w||y<0    ||maze[x][y]==1) return 1;    return 0;}void dfs(wode xx){    int s,t;    if(can_not(xx.x-1,xx.y)    &&can_not(xx.x,xx.y-1)    &&can_not(xx.x+1, xx.y)    &&can_not(xx.x,xx.y+1)) return;        for(int j = 1;j<21;j++)        {    //      if(check == 1) break;            s = xx.x;            t = xx.y+j;            wode xixi;            if(maze[s][t] == 1&&abs(j)==1)                break;            xixi.time = xx.time;            xixi.x=s;            xixi.y=t;            if(s==ex&&t==ey)            {                ans = min(ans, xixi.time);//              debug                break;            }            if(!(s>=0&&s<h&&t>=0&&t<w)) break;            if(s>=0&&s<h&&t>=0&&t<w            &&maze[s][t] == 1&&abs(j)!=1)            {                maze[s][t] = 0;                xixi.x-=0;                xixi.y-=1;                if(xixi.x==ex&&xixi.y==ey)                {                    ans = min(ans, xixi.time);                    return;                }                xixi.time++;            if(xixi.time>9)                    {                        maze[s][t] = 1;                        break;                    }                dfs(xixi);                maze[s][t] = 1;                break;            }        }            for(int j = 1;j<21 ;j++)        {    //      if(check == 1) break;            s = xx.x;            t = xx.y-j;            if(maze[s][t] == 1&&abs(j)==1)                break;            wode xixi;            xixi.time = xx.time;            xixi.x=s;            xixi.y=t;            if(s==ex&&t==ey)            {                ans = min(ans, xixi.time);                break;            }            if(!(s>=0&&s<h&&t>=0&&t<w)) break;            if(s>=0&&s<h&&t>=0&&t<w            &&maze[s][t] == 1&&abs(j)!=1)            {                maze[s][t] = 0;                xixi.x-=0;                xixi.y+=1;                if(xixi.x==ex&&xixi.y==ey                )                {                    ans = min(ans, xixi.time);                    return;                }                xixi.time++;                if(xixi.time>9)                    {                        maze[s][t] = 1;                        break;                    }                dfs(xixi);                maze[s][t] = 1;                break;            }        }            for(int j = 1;j<21 ;j++)        {    //      if(check == 1) break;            s = xx.x+j;            t = xx.y;            if(maze[s][t] == 1&&abs(j)==1)                break;            wode xixi;            xixi.time = xx.time;            xixi.x=s;            xixi.y=t;            if(s==ex&&t==ey)            {                ans = min(ans, xixi.time);                break;            }            if(!(s>=0&&s<h&&t>=0&&t<w)) break;            if(s>=0&&s<h&&t>=0&&t<w            &&maze[s][t] == 1&&abs(j)!=1)            {                maze[s][t] = 0;                xixi.x-=1;                xixi.y-=0;                if(xixi.x==ex&&xixi.y==ey                )                {                    ans = min(ans, xixi.time);                    return;                }                xixi.time++;                if(xixi.time>9)                    {                        maze[s][t] = 1;                        break;                    }                dfs(xixi);                maze[s][t] = 1;                break;            }        }            for(int j = 1;j<21 ;j++)        {    //      if(check == 1) break;            s = xx.x-j;            t = xx.y;            if(maze[s][t] == 1&&abs(j)==1)                break;            wode xixi;            xixi.time = xx.time;            xixi.x=s;            xixi.y=t;            if(s==ex&&t==ey)            {                ans = min(ans, xixi.time);                break;            }            if(!(s>=0&&s<h&&t>=0&&t<w)) break;            if(s>=0&&s<h&&t>=0&&t<w            &&maze[s][t] == 1&&abs(j)!=1)            {                maze[s][t] = 0;                xixi.x+=1;                xixi.y-=0;                if(xixi.x==ex&&xixi.y==ey                )                {                    ans = min(ans, xixi.time);                    return;                }                xixi.time++;                    if(xixi.time>9)                    {                        maze[s][t] = 1;                        break;                    }                dfs(xixi);                maze[s][t] = 1;                break;            }        }}int main(){    while(~scanf("%d%d",&w, &h), w, h)    {        for(int i = 0; i < h; i++)            for(int j = 0; j < w; j++)            {                scanf("%d", &maze[i][j]);                if(maze[i][j] == 2)                    sx = i,sy = j;                else if(maze[i][j]==3)                    ex = i,ey = j;            }        wode mimi;        mimi.time=0;        mimi.x=sx;        mimi.y=sy;dir = 0;        flag = 0; ans = INT_MAX;        dfs(mimi);        if(ans > 9)            puts("-1");        else printf("%d\n", ans+1);    }}

阅读全文

0 0