图的遍历

weixin_30810583

于 2018-11-02 22:22:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/peppa/p/9898818.html

传送门：https://www.luogu.org/problemnew/show/P3916

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 1e5+1;
inline int read()
{
    static char ch;
    while((ch = getchar()) < '0' || ch > '9');
    int ret = ch - 48;
    while((ch = getchar()) >= '0' && ch <= '9')
        ret = ret * 10 + ch - 48;
    return ret;
 }
int n,m,u,v;
vector<int>map[N];
int b[N];
void dfs(int x,int y)
{
    if(b[x] > 0) return;
    b[x] = y;
    for(int i = 0 ;i < map[x].size();i++)
        dfs(map[x][i],y);//所有与x相连的结点所能到达的最大编号都是x 
}
int main()
{
    n = read(); m = read();
    for(int i = 1;i <= m;i++)
    {
        u = read(); v = read();
        map[v].push_back(u);//map连接v->u ,反向建边 
    }
    for(int i = n;i >= 1;i--) dfs(i,i);//i逆序保证与i相连的点所能到达的最大编号都是i 
    for(int i = 1;i <= n;i++) printf("%d ",b[i]);
    return 0;
}