边双联通问题求解(构造边双连通图)POJ3352(Road Construction)

题目链接:传送门

题目大意:给你一副无向图,问至少加多少条边使图成为边双联通图

题目思路:tarjan算法加缩点,缩点后求出度数为1的叶子节点个数,需要加边数为(leaf+1)/2

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <stack>
 8 #include <cctype>
 9 #include <queue>
10 #include <string>
11 #include <vector>
12 #include <set>
13 #include <map>
14 #include <climits>
15 #define lson root<<1,l,mid
16 #define rson root<<1|1,mid+1,r
17 #define fi first
18 #define se second
19 #define ping(x,y) ((x-y)*(x-y))
20 #define mst(x,y) memset(x,y,sizeof(x))
21 #define mcp(x,y) memcpy(x,y,sizeof(y))
22 #define Min(x,y) (x<y?x:y)
23 #define Max(x,y) (x>y?x:y)
24 using namespace std;
25 #define gamma 0.5772156649015328606065120
26 #define MOD 1000000007
27 #define inf 0x3f3f3f3f
28 #define N 5005
29 #define maxn 10005
30 typedef long long LL;
31 typedef pair<int,int> PII;
32 
33 vector<int>V[1005];
34 int n,m,hcnt,deep;
35 struct Node{
36     int to,next;
37     Node(){}
38     Node(int a,int b):to(a),next(b){}
39 }node[N];
40 int head[N],vis[N],low[N],dfn[N];
41 int d[N];
42 
43 inline void add(int x,int y){
44     node[hcnt]=Node(y,head[x]);
45     head[x]=hcnt++;
46 }
47 
48 inline void init(){
49     mst(d,0);
50     mst(vis,0); vis[1]=1;
51     mst(head,-1);
52     hcnt=0;
53     deep=1;
54     low[1]=dfn[1]=1;
55 }
56 
57 void dfs(int x,int father){
58     for(int i=head[x];~i;i=node[i].next){
59         int e=node[i].to;
60         if(e==father)continue;
61         if(vis[e])low[x]=min(low[x],dfn[e]);
62         else{
63             vis[e]=1;
64             low[e]=dfn[e]=++deep;
65             dfs(e,x);
66             low[x]=min(low[x],low[e]);
67         }
68     }
69 }
70 
71 int main(){
72     int i,j,group,Case=0,x,y;
73     while(scanf("%d%d",&n,&m)!=EOF){
74         for(i=0;i<=n;++i)V[i].clear();
75         init();
76         while(m--){
77             scanf("%d%d",&x,&y);
78             add(x,y);
79             add(y,x);
80             V[x].push_back(y);     ///两点之间联通
81             V[y].push_back(x);     ///在缩点之后用于计算度数
82         }
83         dfs(1,-1);    ///tarjan算法
84         int leaf=0;
85         for(i=1;i<=n;++i)
86         if(V[i].size())
87         for(j=0;j<V[i].size();++j){
88             int e=V[i][j];
89             if(low[i]!=low[e])
90                 d[low[i]]++;
91         }
92         for(i=0;i<=deep;++i)
93             if(d[i]==1)++leaf;
94         printf("%d
",(leaf+1)/2);
95     }
96     return 0;
97 }
原文地址:https://www.cnblogs.com/Kurokey/p/5524657.html