标签:back 反向 show ret void turn new main getc
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; }
标签:back 反向 show ret void turn new main getc
原文地址:https://www.cnblogs.com/peppa/p/9898818.html