1，边双联通(桥)
POJ3177Redundant Paths
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17989 | Accepted: 7470 |
Description
In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.
Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.
There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.
Input
Line 1: Two space-separated integers: F and R
Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.
Output
Line 1: A single integer that is the number of new paths that must be built.
Sample Input
7 7 1 2 2 3 3 4 2 5 4 5 5 6 5 7
Sample Output
2
Hint
Explanation of the sample:
One visualization of the paths is:
1 2 3 +---+---+ | | | | 6 +---+---+ 4 / 5 / / 7 +
Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
1 2 3 +---+---+ : | | : | | 6 +---+---+ 4 / 5 : / : / : 7 + - - - -
Check some of the routes:
1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2
1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4
3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7
Every pair of fields is, in fact, connected by two routes.
It‘s possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.
1 /**
2 在树中至少添加多少条边能使图变为双连通图。
3 结论：添加边数=（树中度为1的节点数+1）/2
4
5 具体方法为，首先把两个最近公共祖先最远的两个叶节点之间连接一条边，
6 这样可以把这两个点到祖先的路径上所有点收缩到一起，因为一个形成的
7 环一定是双连通的。然后再找两个最近公共祖先最远的两个叶节点，这样
8 一对一对找完，恰好是(leaf+1)/2次，把所有点收缩到了一起。
9
10 */
11 #include <cstdio>
12 #include <iostream>
13 #include <cstring>
14 #include <algorithm>
15 #include <cmath>
16 #include <stack>
17
18 using namespace std;
19 const int MAXN = 5e3 + 7;
20 const int MAXM = 1e4 + 7;
21
22 struct Edge {
23 int to, w, next;
24 int cut;
25 } edge[MAXM * 2];
26
27 int first[MAXN], dfn[MAXN], low[MAXN], belong[MAXN], degree[MAXN], ins[MAXM];
28
29 stack<int>st;
30
31 int bridge; ///桥的数目
32
33 int sign, n, m;
34
35 int indx, scc;
36
37 inline void init() {
38 memset(first, -1, sizeof(first));
39 sign = 0;
40 }
41
42 inline void add_edge(int u, int v, int w) {
43 edge[sign].to = v;
44 edge[sign].w = w;
45 edge[sign].cut = 0;
46 edge[sign].next = first[u];
47 first[u] = sign++;
48 }
49
50 void tarjan(int now, int pre) {
51 dfn[now] = low[now] = ++indx;
52 st.push(now);
53 ins[now] = 1;
54 for(int i = first[now]; ~i; i = edge[i].next) {
55 int to = edge[i].to;
56 if(to == pre) {
57 continue;
58 }
59 if(!dfn[to]) {
60 tarjan(to, now);
61 low[now] = min(low[now], low[to]);
62 if(low[to] > dfn[now]) {
63 bridge++;
64 edge[i].cut = 1;
65 edge[i^1].cut = 1;
66 }
67 } else if(ins[to]) {
68 low[now] = min(low[now], dfn[to]);
69 }
70 }
71 if(dfn[now] == low[now]) {
72 scc++;
73 int top;
74 do {
75 top = st.top();
76 st.pop();
77 ins[top] = 0;
78 belong[top] = scc;
79 } while(top != now);
80 }
81 }
82
83 int main()
84 {
85 int u, v;
86 while(~scanf("%d %d", &n, &m)) {
87 init();
88 memset(dfn, 0, sizeof(dfn));
89 memset(ins, 0, sizeof(ins));
90 memset(belong, 0, sizeof(belong));
91 memset(degree, 0, sizeof(degree));
92 indx = scc = 0;
93 while(!st.empty()) {
94 st.pop();
95 }
96 for(int i = 1; i<= m; i++ ) {
97 scanf("%d %d", &u, &v);
98 add_edge(u, v, 1);
99 add_edge(v, u, 1);
100 }
101
102 tarjan(1, -1);
103
104 for(int i = 1; i <= n; i++ ) {
105 for(int j = first[i]; ~j; j = edge[j].next) {
106 int from = i, to = edge[j].to;
107 if(edge[j].cut) {
108 degree[ belong[i] ]++;
109 }
110 }
111 }
112 int ans = 0;
113 for(int i = 1; i <= scc; i++ ) {
114 if(degree[i] == 1) {
115 ans ++;
116 }
117 }
118 ans = (ans + 1) / 2;
119 printf("%d\n", ans);
120 }
121
122 return 0;
123 }
2，点双联通
P3388 【模板】割点（割顶）
题目背景
割点
题目描述
给出一个n个点，m条边的无向图，求图的割点。
输入输出格式
输入格式：
第一行输入n,m
下面m行每行输入x,y表示x到y有一条边
输出格式：
第一行输出割点个数
第二行按照节点编号从小到大输出节点，用空格隔开
输入输出样例
输入样例#1：
6 7 1 2 1 3 1 4 2 5 3 5 4 5 5 6
输出样例#1：
1 5
说明
n,m均为100000
tarjan 图不一定联通！！！
注意这题实际数据范围比题目描述要大。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4 const int MAXN = 1e5 + 7;
5 const int MAXM = 2e5 + 7;
6
7 int n, m, first[MAXN], sign, indx, scc;
8
9 int dfn[MAXN], low[MAXN], ins[MAXN], cut[MAXN];
10
11 //stack<int>stk;
12
13 struct Edge {
14 int to, w, next;
15 } edge[MAXM];
16
17 inline void init() {
18 // for(int i = 0; i <= n; i++ ) {
19 // first[i] = -1;
20 // }
21 memset(first, -1, sizeof(first));
22 sign = 0;
23 }
24
25 inline void add_edge(int u, int v, int w) {
26 edge[sign].to = v;
27 edge[sign].w = w;
28 edge[sign].next = first[u];
29 first[u] = sign++;
30 }
31
32 inline void init_tarjan() {
33 for(int i = 0; i <= n; i++ ) {
34 dfn[i] = low[i] = ins[i] = cut[i] = 0;
35 }
36 indx = scc = 0;
37 //while(!stk.empty()) { stk.pop(); }
38 }
39
40 void tarjan(int now, int faz) {
41 low[now] = dfn[now] = ++indx;
42 int child = 0;
43 for(int i = first[now]; ~i; i = edge[i].next) {
44 int to = edge[i].to;
45 if(!dfn[to]) {
46 tarjan(to, faz);
47 low[now] = min(low[now], low[to]);
48 ///非根节点，如果dfn[to] <= low[to]那么它是割点
49 if(low[to] >= dfn[now] && now != faz) {
50 cut[now] = 1;
51 }
52 if(now == faz) {
53 child++;
54 }
55 }
56 low[now] = min(low[now], dfn[to]);
57 }
58 if(child >= 2 && now == faz) { ///根节点有两个以上子树的是割点
59 cut[now] = 1;
60 }
61 }
62
63 int main()
64 {
65 //freopen("testdata.in", "r", stdin);
66 while(~scanf("%d %d", &n, &m)) {
67 init();
68 init_tarjan();
69 for(int i = 1; i <= m; i++ ) {
70 int u, v;
71 scanf("%d %d", &u, &v);
72 add_edge(u, v, 1);
73 add_edge(v, u, 1);
74 }
75 for(int i = 1; i <= n; i++ ) {
76 if(!dfn[i]) {
77 tarjan(i, i);
78 }
79 }
80 vector<int>vec;
81 for(int i = 1; i <= n; i++ ) {
82 if(cut[i]) {
83 vec.push_back(i);
84 }
85 }
86 printf("%d\n", vec.size());
87 for(int i = 0; i < vec.size(); i++ ) {
88 printf("%d ", vec[i]);
89 }
90 }
91 return 0;
92 }
原文地址：https://www.cnblogs.com/Q1143316492/p/9195174.html