HDU 4612--warm up "undirected graph Edge Double Connectivity Bridge number && reconstruction graph to find the diameter of the tree after the contraction point"

Warm up Time limit:10000/5000 MS (java/others) Memory limit:65535/65535 K (java/others)
Total submission (s): 4830 Accepted Submission (s): 1086


Problem Description N Planets is connected by M bidirectional channels the Allow instant transportation. It's always possible to travel between any and planets through these channels.
If We can isolate some planets from others to breaking only one channel, the channel is called a Bridge of the Transporta tion System.
People don ' t like is isolated. So they ask what ' s the minimal number of bridges they can has if they decide to build a new channel.
Note that there could is more than one channel between and one planets.

Input the input contains multiple cases.
Each case starts with positive integers N and M, indicating the number of planets and the number of channels.
(2<=n<=200000, 1<=m<=1000000)
Next M lines each contains-positive integers a and B, indicating a channel between planet A and B in the system. Planets is numbered by 1..N.
A line with the integers ' 0 ' terminates the input.
Output for each case, output the minimal number of bridges after building a new channel in a line.
Sample Input

Sample Output

0

Test instructions: There are n points, M-Edge (there is a heavy edge) of the graph, so that there may be a bridge in the picture, asked to add an edge, so that the bridge to the minimum, to seek the bridge tree.

Idea: Figure out the number of bridges in the diagram, and then the reconstruction map must be a tree, the longest diameter of the tree, in the diameter of the two ends of the point connected, the figure of the bridge will be the least.

#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack > #define MAXN 200000+1000#define MAXM 2000000+1000#pragma COMMENT (linker, "/stack:1024000000,1024000000") using Namespace Std;int N, m;struct node {int u, V, next;}; Node Edge[maxm];int HEAD[MAXN], Cnt;int DFN[MAXN], Low[maxn];int STACK[MAXN], Top;bool instack[maxn];int BELONG[MAXN]; int dfs_clock;int ebc_clock;int bridge;vector<int>map[maxn];int dist[maxn], vis[maxn];int ans, nod;void init () {C    NT = 0; Memset (Head,-1, sizeof (head));}    void Addedge (int u, int v) {edge[cnt] = {u, V, Head[u]}; Head[u] = cnt++;}        void Getmap () {while (m--) {int A, B;        scanf ("%d%d", &a, &b);        Addedge (A, b);    Addedge (b, a);    }}void Tarjan (int u, int per) {int V;    Low[u] = dfn[u] = ++dfs_clock;    stack[top++] = u;    Instack[u] = true;    int has = 1;        for (int i = head[u]; I! =-1; i = edge[i].next) {v = edge[i].v; if (V= = per && have) {has = 0;        Continue            } if (!dfn[v]) {Tarjan (V, u);            Low[u] = min (Low[u], low[v]);        if (Low[v] > Dfn[u]) bridge++;    } else if (Instack[v]) low[u] = min (Low[u], dfn[v]);        } if (dfn[u] = = Low[u]) {ebc_clock++;            do{v = stack[--top];            INSTACK[V] = false;        BELONG[V] = Ebc_clock;    }while (v! = u);    }}void Suodian () {//ëõµã½¨ðâí¼for (int i = 1; I <= ebc_clock; ++i) map[i].clear ();        for (int i = 0; i < cnt; i = i + 2) {int u = belong[edge[i].u];        int v = belong[edge[i].v];            if (U = v) {map[u].push_back (v);        Map[v].push_back (U);    }}}void Find () {memset (low, 0, sizeof (low));    memset (DFN, 0, sizeof (DFN));    Memset (Instack, False, sizeof (Instack));    memset (Belong, 0, sizeof (Belong));    Dfs_clock = Ebc_clock = top = Bridge = 0; for (int i =1; I <= N;    ++i) {if (!dfn[i]) Tarjan (i, I);    }}void BFS (int x) {queue<int>q;    memset (Vis, 0, sizeof (VIS));    memset (Dist, 0, sizeof (Dist));    VIS[X] = 1;    Ans = 0;    nod = x;    Q.push (x);        while (!q.empty ()) {int u = q.front ();        Q.pop ();            for (int i = 0; i < map[u].size (); ++i) {int v = map[u][i];                if (!vis[v]) {vis[v] = 1;                DIST[V] = Dist[u] + 1;                    if (ans < dist[v]) {nod = V;                ans = dist[v];            } q.push (v);        }}}}int Main () {while (scanf ("%d%d", &n, &m), n | | m) {init ();        Getmap ();        Find ();        Suodian ();        BFS (1);        BFS (NOD);    printf ("%d\n", Bridge-ans); } return 0;}

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HDU 4612--warm up "undirected graph Edge Double Connectivity Bridge number && reconstruction graph to find the diameter of the tree after the contraction point"