hdu 4612 Warm up

题目大意，给一个图，加一条边使图中剩余的桥最少，求最少桥的数量。

先缩点，得到一棵树，求树的直径，答案=桥-直径。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <vector>#include <stack>#include <queue>#include <map>#include <algorithm>#include <iostream>#include <string>#include <set>#define X first#define Y second#define sqr(x) (x)*(x)using namespace std;const double PI = acos(-1.0);map<int,int>::iterator it;typedef long long LL ;#pragma comment(linker, "/STACK:102400000,102400000")const int N = 200002;const int M = 2000003*2;struct edge{    int v,pre;} E[M];int head[N],cnt;int belong[N],qcnt,t_stamp,low[N],dfn[N];bool vis[N],instack[N];stack<int> SK;vector<int> QLT[N];int cnt_brigde;int mxlen,top;void init(){    cnt_brigde=0;    for(int i=0; i<=200000; ++i)    {        QLT[i].clear();    }    mxlen=0;    cnt=0;    qcnt=0;    t_stamp=0;    top=0;    while(!SK.empty())SK.pop();    memset(dfn,-1,sizeof(dfn));    memset(head,-1,sizeof(head));    memset(instack,0,sizeof(instack));    memset(vis,false,sizeof(vis));}void add_edge(int u,int v){    E[cnt].v=v;    E[cnt].pre=head[u];    head[u]=cnt++;}int n,m;int stk[N*2];void dfs1(int u,int pre){    //printf("%d   %d\n",u,pre);    low[u]=dfn[u]=++t_stamp;    stk[++top]=u;    instack[u]=true;    int v,cntpre=0;    for(int e=head[u]; ~e; e=E[e].pre)    {        v = E[e].v;        if(v==pre&&cntpre==0)        {            ++cntpre;            continue;        }        //printf("to %d\n",v);        if(dfn[v]==-1)        {            dfs1(v,u);            low[u]=min(low[u],low[v]);            if(dfn[u]<low[v])            {                ++cnt_brigde;            }        }        else if(instack[v]&&low[u]>dfn[v])        {            low[u]=dfn[v];        }    }    //puts("hered");    if(low[u]==dfn[u])    {        while(stk[top]!=u)        {            v = stk[top--];            belong[v]=qcnt;            QLT[qcnt].push_back(v);        }        QLT[qcnt].push_back(u);        belong[u]=qcnt++;        --top;    }}int mxdep,dpstp;int dp[N][2];void dfs2(int u,int pre){    dp[u][1]=dp[u][0]=0;    vis[u]=true;    int v;    //printf("qlt %d:\n",u);    for(int i=0; i<QLT[u].size(); ++i)    {        for(int e=head[QLT[u][i]]; ~e; e=E[e].pre)        {            //printf("v = %d\n",v);            v = belong[E[e].v];            if(!vis[v]&&v!=pre&&v!=u)            {                dfs2(v,u);                int vmx = dp[v][0]+1;                if(dp[u][0]<vmx)                {                    dp[u][1]=dp[u][0];                    dp[u][0]=vmx;                }                else if(dp[u][1]<vmx)                {                    dp[u][1]=vmx;                }            }        }    }    mxlen = max(mxlen,dp[u][0]+dp[u][1]);}int main(){    while(true)    {        scanf("%d%d",&n,&m);        if(n==0&&m==0)break;        init();        int u,v;        for(int i=0; i<m; ++i)        {            scanf("%d%d",&u,&v);            add_edge(u,v);            add_edge(v,u);        }        dfs1(1,1);        dfs2(0,0);        //printf("  %d %d\n",cnt_brigde,mxlen);        printf("%d\n",cnt_brigde-mxlen);    }    return 0;}