CSUOJ 1010 Water Drinking（BFS）

1010: Water Drinking

Description

The Happy Desert is full of sands. There is only a kind of animal called camel living on the Happy Desert. ‘Cause they live here, they need water here. Fortunately, they find a pond which is full of water in the east corner of the desert. Though small, but enough. However, they still need to stand in lines to drink the water in the pond.

Now we mark the pond with number 0, and each of the camels with a specific number, starting from 1. And we use a pair of number to show the two adjacent camels, in which the former number is closer to the pond. There may be more than one lines starting from the pond and ending in a certain camel, but each line is always straight and has no forks.

Input

There’re multiple test cases. In each case, there is a number N (1≤N≤100000) followed by N lines of number pairs presenting the proximate two camels. There are 99999 camels at most.

Output

For each test case, output the camel’s number who is standing in the last position of its line but is closest to the pond. If there are more than one answer, output the one with the minimal number.

Sample Input

1

0 1

5

0 2

0 1

1 4

2 3

3 5

5

1 3

0 2

0 1

0 4

4 5

Sample Output

1

4

2

题目大意：在一个沙漠中有一个水池，一些骆驼在排队喝水。骆驼排成了很多队伍，队伍的起点都是水池，求排在队伍最后但是离水池最近的骆驼编号。

解题思路：建图然后BFS，用一个数组记录某个骆驼是否在队尾，最后比较距离，输出距离最近且编号最小的骆驼即可。

代码如下：

#include <bits/stdc++.h>#define INF 1e9using namespace std;const int maxn = 100005;int head[maxn],cnt,d[maxn],n;bool en[maxn];struct Edge{    int to,next;}edge[maxn];void init(){    cnt = 0;    memset(head,-1,sizeof(head));    memset(en,0,sizeof(en));}void add_edge(int from,int to){    edge[cnt].to = to;    edge[cnt].next = head[from];    head[from] = cnt++;}queue<int> que;void bfs(){    while(que.size()) que.pop();    for(int i = 0;i <= n;i++) d[i] = INF;    d[0] = 0;    que.push(0);    while(que.size()){        int u = que.front();        que.pop();        if(head[u] == -1){            en[u] = true;            continue;        }        for(int i = head[u];~i;i = edge[i].next){            int v = edge[i].to;            d[v] = d[u] + 1;            que.push(v);        }    }    return ;}int main(void){    int t,a,b;    while(cin >> n){        init();        for(int i = 0;i < n;i++){            cin >> a >> b;            add_edge(a,b);        }        bfs();        int minVal = INF;        int pos = INF;        for(int i = 1;i <= n;i++){            if(en[i] && minVal > d[i]){                minVal = d[i];                pos = i;            }        }        cout << pos << endl;    }    return 0;}