BZOJ5203 [NEERC2017 Northern] Grand Test 【dfs树】【构造】

题目分析：

　　首先观察可知这是一个无向图，那么我们构建出它的dfs树。由于无向图的性质我们可以知道它的dfs树只有返祖边。考虑下面这样一个结论。

　　结论：若一个点的子树中(包含自己)有两个点有到它祖先的返祖边(不包括到它自己)，

　　首先我们证明S和T肯定在DFS树中是祖先关系，接着证明到T至少有一条返祖边，那么这个结论就是显然的了。

代码：

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3
  4 const int maxn = 102000;
  5
  6 int n,m;
  7 vector <int>g[maxn];
  8
  9 int up[maxn],dep[maxn],wh[maxn],fa[maxn];
 10 int flag, from, to;
 11
 12 void read(){
 13     scanf("%d%d",&n,&m);
 14     for(int i=1;i<=n;i++) g[i].clear();
 15     for(int i=1;i<=m;i++){
 16     int u,v; scanf("%d%d",&u,&v);
 17     g[u].push_back(v); g[v].push_back(u);
 18     }
 19 }
 20
 21 int r1[maxn],num1;
 22 int r2[maxn],num2;
 23 int r3[maxn],num3;
 24 int md;
 25 stack <int> s;
 26 void print(int now){
 27     int pla = now;
 28     while(dep[pla] != to) r1[++num1] = pla,pla = fa[pla];
 29     r1[++num1] = pla;
 30
 31     pla = md;
 32     while(dep[pla] != dep[now]) s.push(pla),pla = fa[pla];
 33     s.push(pla);
 34     while(!s.empty()){r2[++num2] = s.top();s.pop();}
 35     r2[++num2] = r1[num1];
 36
 37     pla = wh[now];
 38     while(dep[pla] != dep[now]) s.push(pla),pla = fa[pla];
 39     s.push(pla);
 40     while(!s.empty()){r3[++num3] = s.top();s.pop();}
 41     pla = r1[num1];
 42     while(dep[pla] != up[now]) s.push(pla),pla = fa[pla];
 43     s.push(pla);
 44     while(!s.empty()){r3[++num3] = s.top();s.pop();}
 45
 46     printf("%d %d\n",r1[1],r1[num1]);
 47     printf("%d ",num1);for(int i=1;i<=num1;i++) printf("%d ",r1[i]);
 48     puts("");
 49     printf("%d ",num2);for(int i=1;i<=num2;i++) printf("%d ",r2[i]);
 50     puts("");
 51     printf("%d ",num3);for(int i=1;i<=num3;i++) printf("%d ",r3[i]);
 52     puts("");flag = 1;
 53 }
 54
 55 void dfs(int now,int dp){
 56     dep[now] = dp;
 57     for(int i=0;i<g[now].size();i++){
 58     int nxt = g[now][i];
 59     if(fa[now] == nxt) continue;
 60     if(flag) break;
 61     if(dep[nxt] == 0){
 62         fa[nxt] = now;
 63         dfs(nxt,dp+1); if(flag) break;
 64         if(up[nxt] >=  dep[now]) continue;
 65         if(up[now] >= 10000000) up[now] = up[nxt],wh[now] = wh[nxt];
 66         else{
 67         if(up[now] <= up[nxt]){
 68             from = now; to = up[nxt];md = wh[nxt];
 69         }else {
 70             from = now; to = up[now];up[now] = up[nxt];
 71             md = wh[now];wh[now]= wh[nxt];
 72         }
 73         print(from);
 74         }
 75     }else{
 76         if(dep[nxt] > dep[now]) continue;
 77         if(up[now] >= 10000000) up[now] = dep[nxt],wh[now] = now;
 78         else{
 79         if(up[now] <= dep[nxt]){
 80             from = now; to = dep[nxt];md = now;
 81         }else {
 82             from = now; to = up[now];up[now] = dep[nxt];
 83             md = wh[now];wh[now] = now;
 84         }
 85         print(from);
 86         }
 87     }
 88     }
 89 }
 90
 91 void work(){
 92     flag = 0; from = 0; to = 0; num1 = num2 = num3 = 0;
 93     for(int i=0;i<=n;i++) wh[i] = 0,fa[i] = 0,dep[i] = 0,up[i] = 1e9;
 94     for(int i=1;i<=n;i++) if(!dep[i]) dfs(i,1);
 95     if(!flag){puts("-1");}
 96 }
 97
 98 int main(){
 99     int cas; scanf("%d",&cas);
100     while(cas--){
101     read();
102     work();
103     }
104     return 0;
105 }

原文地址：https://www.cnblogs.com/Menhera/p/8900620.html