hdu 2460 Network (雙連通分支+暴力LCA)

題意：在一張圖中給出q個加邊操做，問你每次操做以後圖中割邊的個數。點數1e5詢問1000

思路：這道題的作法是先對圖進行縮點，而後變成一顆樹，每次添加新邊如果邊的兩個端點屬於不一樣的分支則必定會造成一個環，這時暴力lca標記全部換上的邊有割邊變爲不是割邊。每次統計就能夠了。理論上說，每次給V字形的圖複雜度會是1e8還有多組數據可能會超時，但這道題能經過。另外這道題會爆棧，注意c++擴棧。

代碼以下：

  1 /**************************************************
  2  * Author     : xiaohao Z
  3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
  4  * Last modified : 2014-09-01 15:25
  5  * Filename     : hdu_2460.cpp
  6  * Description     : 
  7  * ************************************************/
  8 #pragma comment(linker, "/STACK:1024000000,1024000000")
  9 #include <iostream>
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <cstdlib>
 13 #include <string>
 14 #include <cmath>
 15 #include <algorithm>
 16 #include <queue>
 17 #include <stack>
 18 #include <vector>
 19 #include <set>
 20 #include <map>
 21 #define MP(a, b) make_pair(a, b)
 22 #define PB(a) push_back(a)
 23 
 24 using namespace std;
 25 typedef long long ll;
 26 typedef pair<int, int> pii;
 27 typedef pair<unsigned int,unsigned int> puu;
 28 typedef pair<int, double> pid;
 29 typedef pair<ll, int> pli;
 30 typedef pair<int, ll> pil;
 31 
 32 const int INF = 0x3f3f3f3f;
 33 const double eps = 1E-6;
 34 const int LEN = 100010;
 35 vector<int> Map[LEN], tMap[LEN];
 36 stack<int> s;
 37 int n, m, dclock, low[LEN], dfn[LEN], block, blong[LEN];
 38 int dep[LEN], parent[LEN], is[LEN];
 39 
 40 void init()
 41 {
 42     memset(dfn, -1, sizeof dfn);
 43     memset(is, 0, sizeof is);
 44     memset(parent, 0, sizeof parent);
 45     dclock = is[1] = 1; block = 0;
 46     while(!s.empty())s.pop();
 47     for(int i=0; i<LEN; i++) Map[i].clear();
 48 }
 49 
 50 void dfs(int u, int fa){
 51     int v, flag = 0;
 52     dfn[u] = low[u] = dclock++;
 53     s.push(u);
 54     for(int i=0; i<Map[u].size(); i++){
 55         v = Map[u][i];
 56         if(v==fa && !flag){flag = 1;continue;}
 57         if(dfn[v]<0){
 58             dfs(v, u);
 59             low[u] = min(low[u], low[v]);
 60         }else if(low[u]>low[v]) low[u] = min(low[u], dfn[v]);
 61     }
 62     if(low[u] == dfn[u]){
 63         block ++;
 64         do{
 65             v = s.top(); s.pop();
 66             blong[v] = block;
 67         }while(v!=u);
 68     }
 69 }
 70 
 71 void DFS(int v, int fa, int d){
 72     dep[v] = d;
 73     parent[v] = fa;
 74     for(int i=0; i<tMap[v].size(); i++){
 75         int x = tMap[v][i];
 76         if(x != fa)    DFS(x, v, d+1);
 77     }
 78 }
 79 
 80 void build(){
 81     int a, b;
 82     map<pii, int> mp; mp.clear();
 83     for(int i=0; i<LEN; i++) tMap[i].clear();
 84     for(int i=0; i<n; i++)
 85         for(int j=0; j<Map[i].size(); j++){
 86             int x = Map[i][j];
 87             if(blong[i] == blong[x]) continue;
 88             a = blong[i], b = blong[x];
 89             if(mp.count(MP(a, b))) continue;
 90             mp[MP(a, b)] = mp[MP(b, a)] = 1;
 91             tMap[a].PB(b); tMap[b].PB(a);
 92         }
 93     DFS(1, 0, 0);
 94 }
 95 
 96 void _(int val) {cout << "\"" << val << endl;}
 97 
 98 
 99 void lca(int a, int b){
100     if(dep[a] < dep[b]) swap(a, b);
101     int tmp = dep[a] - dep[b];
102     for(int i=0; i<tmp; i++) {
103         if(!is[a]) {is[a] = 1;block --;}
104         a = parent[a];
105     }
106     while(a != b) {
107         if(!is[a]) {is[a] = 1; block --;}
108         if(!is[b]) {is[b] = 1; block --;}
109         a = parent[a], b = parent[b];
110     }
111 }
112 
113 int main()
114 {
115 //    freopen("in.txt", "r", stdin);
116 
117     int a, b, q, kase = 1;
118     while(scanf("%d%d", &n, &m) != EOF){
119         if(!n && !m) break;
120         init();
121         for(int i=0; i<m; i++){
122             scanf("%d%d", &a, &b);
123             a--, b--;
124             Map[a].PB(b); Map[b].PB(a);
125         }
126         dfs(0, -1);
127         build();
128         scanf("%d", &q);
129         printf("Case %d:\n", kase ++);
130         for(int i=0; i<q; i++){
131             scanf("%d%d", &a, &b);
132             a--; b--;
133             if(blong[a] != blong[b]) lca(blong[a], blong[b]);
134             printf("%d\n", block - 1);
135         }
136         puts("");
137     }
138     return 0;
139 }

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