POJ 3352-road Construction (graph theory-bilateral unicom branch algorithm)

The main idea: a graph that requires you to add a minimum of edges, making the last image of a side-double connected branch. The so-called side of the two connected branches, that is, there is no bridge Connectivity branch (the title guarantee data of any two points are connected).

Problem-solving ideas: First use the Tarjan algorithm to set up a DAG map, and then the search degree of 1 points there are number x, then the need to add the edge is (x+1)/2;

At first I wrote this, WA, and then found the following two data, found that can not be too.

#include <stdio.h>#include<Set>#include<vector>#include<string.h>#include<algorithm>using namespacestd;Const intN =1003; Vector<int>G[n];vector<pair<int,int> >DAG;intDfn[n], low[n], mk[n];inttot;intN, M;voidinit () {tot=0;    Dag.clear ();  for(intI=1; i<=n; ++i) {Mk[i]=0;        G[i].clear (); Dfn[i]= Low[i] =-1; }}voidTarjan (intUintf) {Dfn[u]= Low[u] = + +tot;  for(inti =0; I < g[u].size (); ++i) {intv =G[u][i]; if(Dfn[v] = =-1) {Tarjan (V, u); Low[u]=min (Low[u], low[v]); if(Dfn[u] <Low[v])        Dag.push_back (Make_pair (Low[u], low[v])); }        Else if(V! =f) Low[u]=min (Low[u], dfn[v]); }}voidsolve () {init ();  for(intI=1; i<=m; ++i) {intu, v; scanf ("%d%d", &u, &v);        G[u].push_back (v);    G[v].push_back (U); } Tarjan (1, -1);  for(intI=0; I<dag.size (); ++i) {pair<int,int> S =Dag[i]; Mk[s.first]+ +, Mk[s.second] + +; }    intAns =0;  for(intI=1; i<=n; ++i) {if(Mk[i] = =1) ans++; } printf ("%d\n", (ans +1) /2);}intMain () { while(SCANF ("%d%d", &n, &m)! =EOF) solve (); return 0;}

View Code

Special attention needs to be paid to two sets of data:

2 2

1 2

1 2

2 1

1 2

The answers were:

0

1

The code is as follows:

#include <stdio.h>#include<Set>#include<vector>#include<string.h>#include<algorithm>using namespacestd;Const intN =1003; Vector<int>G[n];intDfn[n], low[n], mk[n];inttot;intN, M;voidinit () {tot=0;  for(intI=1; i<=n; ++i) {Mk[i]=0;        G[i].clear (); Dfn[i]= Low[i] =-1; }}voidTarjan (intUintf) {Dfn[u]= Low[u] = + +tot;  for(inti =0; I < g[u].size (); ++i) {intv =G[u][i]; if(Dfn[v] = =-1) {Tarjan (V, u); Low[u]=min (Low[u], low[v]); }        Else if(V! =f) Low[u]=min (Low[u], dfn[v]); }}voidsolve () {init ();  for(intI=1; i<=m; ++i) {intu, v; scanf ("%d%d", &u, &v);        G[u].push_back (v);    G[v].push_back (U); } Tarjan (1, -1);  for(inti =1; I <= N; ++i) { for(intj =0; J < G[i].size (); ++j) {if(Low[i]! =Low[g[i][j]]) Mk[low[i]++; }    }    intAns =0;  for(inti =1; I <= N; ++i)if(Mk[i] = =1) ans++; printf ("%d\n", (ans +1) /2);}intMain () { while(SCANF ("%d%d", &n, &m)! =EOF) solve (); return 0;}

View Code

POJ 3352-road Construction (graph theory-bilateral unicom branch algorithm)