cf1037D. Valid BFS?(BFS?)
阿新 • • 發佈:2018-10-05
題意 ons print tchar http != namespace read for
題意
題目鏈接
Sol
非常妙的一道題。。
可以這樣想,在BFS序中較早出現的一定是先訪問的,所以把每個點連出去的邊按出現的前後順序排個序
看一下按順序遍歷出來的序列與給出的是否相同就行了
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, a[MAXN], dfn[MAXN], tot, vis[MAXN], tim[MAXN]; vector<int> v[MAXN]; int comp(const int &x, const int &y) { return tim[x] < tim[y]; } void BFS() { queue<int> q; q.push(1); while(!q.empty()) { int p = q.front(); q.pop(); dfn[++tot] = p; vis[p] = 1; for(int i = 0, to; i < v[p].size(); i++) { if(!vis[(to = v[p][i])]) q.push(to); } } } main() { N = read(); for(int i = 1; i <= N - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } for(int i = 1; i <= N; i++) tim[a[i] = read()] = i; for(int i = 1; i <= N; i++) sort(v[i].begin(), v[i].end(), comp); BFS(); //for(int i = 1; i <= N; i++) printf("%d ", dfn[i]); puts(""); for(int i = 1; i <= N; i++) if(dfn[i] != a[i]) {puts("No"); return 0;} puts("Yes"); }
cf1037D. Valid BFS?(BFS?)