国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

POJ3009-Curling 2.0-DFS

2019-11-08 18:28:54
字體:
來源:轉載
供稿:網友

原題鏈接 Curling 2.0 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19909 Accepted: 8136 Description

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will PRoceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.

Fig. 1: Example of board (S: start, G: goal)

The movement of the stone obeys the following rules:

At the beginning, the stone stands still at the start square. The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited. When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)). Once thrown, the stone keeps moving to the same direction until one of the following occurs: The stone hits a block (Fig. 2(b), (c)). The stone stops at the square next to the block it hit. The block disappears. The stone gets out of the board. The game ends in failure. The stone reaches the goal square. The stone stops there and the game ends in success. You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.

Fig. 2: Stone movements

Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.

With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).

Fig. 3: The solution for Fig. D-1 and the final board configuration

Input

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.

Each dataset is formatted as follows.

the width(=w) and the height(=h) of the board First row of the board … h-th row of the board

The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.

Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.

0 vacant square 1 block 2 start position 3 goal position The dataset for Fig. D-1 is as follows:

6 6 1 0 0 2 1 0 1 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1

Output

For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.

Sample Input

2 1 3 2 6 6 1 0 0 2 1 0 1 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 6 1 1 1 2 1 1 3 6 1 1 0 2 1 1 3 12 1 2 0 1 1 1 1 1 1 1 1 1 3 13 1 2 0 1 1 1 1 1 1 1 1 1 1 3 0 0 Sample Output

1 4 -1 4 10 -1 Source

Japan 2006 Domestic

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int n,m,sx,sy,gx,gy,minres,a[22][22];const int dx[4]={0,-1,0,1};const int dy[4]={1,0,-1,0};bool can(int x,int y){ if(x<n && x >-1 && y<m && y > -1 && a[x][y]!=1) return true; else return false;}void dfs(int x,int y,int res){ if(res>10) return; if(x==gx && y==gy){ if(minres==-1){ minres=res; return; } minres=min(res,minres); return; } for(int i=0;i<4;i++){ if(dx[i]==0){ if(!can(x,y+dy[i])) continue;//當前方向必須沒有阻擋 int j; for(j=1;can(x,y+j*dy[i]);j++){//j從1開始防止相隔的這一格就是終點 if(x==gx && y+j*dy[i]==gy){ dfs(gx,gy,res+1); goto END; } } int nx=x,ny=y+j*dy[i]; if(nx<n && nx >-1 && ny<m && ny > -1){//此時已經可以確定就是遇到石頭了 a[nx][ny]=0; dfs(nx,ny-dy[i],res+1); a[nx][ny]=1; } } else{ if(!can(x+dx[i],y)) continue; int j; for(j=1;can(x+j*dx[i],y);j++){ if(x+j*dx[i]==gx && y==gy){ dfs(gx,gy,res+1); goto END; } } int nx=x+j*dx[i],ny=y; if(nx<n && nx >-1 && ny<m && ny > -1){ a[nx][ny]=0; dfs(nx-dx[i],ny,res+1); a[nx][ny]=1; } } END:; } return;}int main(){ while(scanf("%d%d",&m,&n)==2 && n+m != 0){ minres=-1; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ scanf("%d",&a[i][j]); if(a[i][j]==2){ sx=i;sy=j; } else if(a[i][j]==3){ gx=i;gy=j; } } } dfs(sx,sy,0); cout << minres << endl; } return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 承德县| 纳雍县| 白城市| 阿坝| 桂林市| 花垣县| 泗洪县| 慈溪市| 平潭县| 岗巴县| 锦州市| 丰台区| 阳江市| 阜宁县| 鄄城县| 霍州市| 五台县| 康平县| 芦山县| 徐水县| 苏州市| 岳西县| 咸宁市| 郁南县| 民权县| 陈巴尔虎旗| 浙江省| 昆明市| 巴南区| 铁力市| 武宣县| 昌平区| 德令哈市| 涞水县| 开平市| 枣庄市| 大新县| 黑龙江省| 阳城县| 资溪县| 五大连池市|