Codeforces 793B Igor and his way to work （dfs/bfs

最新推荐文章于 2019-07-04 10:28:04 发布

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最新推荐文章于 2019-07-04 10:28:04 发布

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Igor是一名财务分析师，他需要从家出发前往银行上班，但城市部分道路因施工封闭且他的车只能转弯两次。本篇介绍了如何通过算法帮助Igor找到合适的路径。

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B. Igor and his way to work

Description

Woken up by the alarm clock Igor the financial analyst hurried up to the work. He ate his breakfast and sat in his car. Sadly, when he opened his GPS navigator, he found that some of the roads in Bankopolis, the city where he lives, are closed due to road works. Moreover, Igor has some problems with the steering wheel, so he can make no more than two turns on his way to his office in bank.

Bankopolis looks like a grid of n rows and m columns. Igor should find a way from his home to the bank that has no more than two turns and doesn’t contain cells with road works, or determine that it is impossible and he should work from home. A turn is a change in movement direction. Igor’s car can only move to the left, to the right, upwards and downwards. Initially Igor can choose any direction. Igor is still sleepy, so you should help him.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the grid.

Each of the next n lines contains m characters denoting the corresponding row of the grid. The following characters can occur:

“.” — an empty cell;
“*” — a cell with road works;
“S” — the cell where Igor’s home is located;
“T” — the cell where Igor’s office is located.
It is guaranteed that “S” and “T” appear exactly once each.

Output

In the only line print “YES” if there is a path between Igor’s home and Igor’s office with no more than two turns, and “NO” otherwise.

Sample Input

5 5
..S..
****.
T....
****.
.....

5 5
S....
****.
.....
.****
..T..

Sample Output

YES
NO

Hint

The first sample is shown on the following picture: 这里写图片描述
In the second sample it is impossible to reach Igor’s office using less that 4 turns, thus there exists no path using no more than 2 turns. The path using exactly 4 turns is shown on this picture:

题意 :

从一个点到另一个点其中转弯次数不能超过2次

题解：

还不很理解明天补上吧

AC代码

#include <bits/stdc++.h>
using namespace std;
char str[1005][1005];
int vis[1005][5005];
int n, m;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};

void dfs(int x, int y, int d, int pre)
{
    if(x<0||x>=n||y<0||y>=m) return ;
    if(str[x][y] == '*') return ;
    vis[x][y] = 1;
    if(d==1) {
        for(int i = 0; i < 4; i++) dfs(x+dx[i],y+dy[i],0,i);
    }
    dfs(x+dx[pre],y+dy[pre],d,pre);
}
bool ans_dfs(int x,int y,int pre)
{
    if(x<0||x>=n||y<0||y>=m) return false;
    if(str[x][y]=='*') return false;
    if(vis[x][y]==1) return true;
    return ans_dfs(x+dx[pre],y+dy[pre],pre);
}
int main()
{
    memset(vis,0,sizeof(vis));
    int x1,x2,y1,y2;
    scanf("%d%d",&n,&m);
    for(int i = 0;i < n; i++) 
        scanf("%s",str[i]);
    for(int i = 0;i < n; i++) {
        for(int j = 0;j < m; j++) {
            if(str[i][j]=='S') {
                x1 = i;
                y1 = j;
            }
            if(str[i][j]=='T') {
                x2 = i;
                y2 = j;
            }
        }
    }
    for(int i = 0;i < 4; i++) {
        dfs(x1,y1,1,i);
    }
    bool flag = false;
    for(int i = 0;i < 4; i++) {
        if(ans_dfs(x2,y2,i)) {
            flag = true;
            break;
        }
    }
    if(flag) puts("YES");
    else puts("NO");
return 0;
}