sicily1155. Can I Post the lette

1155. Can I Post the lette

 

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

 

I am a traveler. I want to post a letter to Merlin. But because there are so many roads I can walk through, and maybe I can’t go to Merlin’s house following these roads, I must judge whether I can post the letter to Merlin before starting my travel. 

Suppose the cities are numbered from 0 to N-1, I am at city 0, and Merlin is at city N-1. And there are M roads I can walk through, each of which connects two cities. Please note that each road is direct, i.e. a road from A to B does not indicate a road from B to A.

Please help me to find out whether I could go to Merlin’s house or not.

 

Input

 

There are multiple input cases. For one case, first are two lines of two integers N and M, (N<=200, M<=N*N/2), that means the number of citys and the number of roads. And Merlin stands at city N-1. After that, there are M lines. Each line contains two integers i and j, what means that there is a road from city i to city j.

The input is terminated by N=0.

 

Output

For each test case, if I can post the letter print “I can post the letter” in one line, otherwise print “I can't post the letter”.

Sample Input

320 11 2310 10

Sample Output

I can post the letterI can't post the letter

Problem Source

ZSUACM Team Member

 

题目分析：一次搜索，拓扑即可完成。

 

参考代码：

#include <iostream>#include <queue>#include <cstring>using namespace std;struct R{    int beg;    int end;};int main(){    int n, m;    bool vis[200];    while (1)    {        memset(vis, false, sizeof(vis));        cin >> n;        if (n == 0) break;        cin >> m;        R road[m];        for (int i = 0; i < m; ++ i)        {            cin >> road[i].beg >> road[i].end;        }        vis[0] = true;        queue<int> q;        q.push(0);        while (!q.empty() && !vis[n - 1])        {            int temp = q.front();            for (int i = 0; i < m; ++ i)            {                if (temp == road[i].beg)                {                    int tem = road[i].end;                    if (!vis[tem])                    {                        q.push(tem);                        vis[tem] = true;                    }                }            }            q.pop();        }        if (vis[n - 1])        {            cout << "I can post the letter" << endl;        }        else        {            cout << "I can't post the letter" << endl;        }            }}