hdu - 2667 Proving Equivalences(强连通)

http://acm.hdu.edu.cn/showproblem.php?pid=2767

求至少添加多少条边才能变成强连通分量.统计入度为0的点和出度为0的点,取最大值即可.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>

 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("dout.txt", "w", stdout);

 using namespace std;
 #define N 20100
 //N为最大点数
 #define M 50100
 //M为最大边数
 int n, m;//n m 为点数和边数

 struct Edge{
     int from, to, nex;
     bool sign;//是否为桥
 }edge[M<<];
 int head[N], edgenum;
 void add(int u, int v){//边的起点和终点
     Edge E={u, v, head[u], false};
     edge[edgenum] = E;
     head[u] = edgenum++;
 }

 int DFN[N], Low[N], Stack[N], top, Time; //Low[u]是点集{u点及以u点为根的子树} 中(所有反向弧)能指向的(离根最近的祖先v) 的DFN[v]值(即v点时间戳)
 int taj;//连通分支标号，从1开始
 int Belong[N];//Belong[i] 表示i点属于的连通分支
 bool Instack[N];
 vector<int> bcc[N]; //标号从1开始

 void tarjan(int u ,int fa){
     DFN[u] = Low[u] = ++ Time ;
     Stack[top ++ ] = u ;
     Instack[u] =  ;

     for (int i = head[u] ; ~i ; i = edge[i].nex ){
         int v = edge[i].to ;
         if(DFN[v] == -)
         {
             tarjan(v , u) ;
             Low[u] = min(Low[u] ,Low[v]) ;
             if(DFN[u] < Low[v])
             {
                 edge[i].sign = ;//为割桥
             }
         }
         else if(Instack[v]) Low[u] = min(Low[u] ,DFN[v]) ;
     }
     if(Low[u] == DFN[u]){
         int now;
         taj ++ ; bcc[taj].clear();
         do{
             now = Stack[-- top] ;
             Instack[now] =  ;
             Belong [now] = taj ;
             bcc[taj].push_back(now);
         }while(now != u) ;
     }
 }

 void tarjan_init(int all){
     memset(DFN, -, sizeof(DFN));
     memset(Instack, , sizeof(Instack));
     top = Time = taj = ;
     for(int i=;i<=all;i++)if(DFN[i]==- )tarjan(i, i); //注意开始点标！！！
 }
 vector<int>G[N];
 int du[N];
 void suodian(){
     memset(du, , sizeof(du));
     for(int i = ; i <= taj; i++)G[i].clear();
     for(int i = ; i < edgenum; i++){
         int u = Belong[edge[i].from], v = Belong[edge[i].to];
         if(u!=v)
         {
             G[u].push_back(v), du[v]++;
            // printf("%d %d\n",u,v);
         }
     }
 }
 void init(){memset(head, -, sizeof(head)); edgenum=;}
 int main()
 {
     //Read();
     int t,a,b;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         init();
         for(int i=;i<m;i++)
         {
             scanf("%d%d",&a,&b);
             add(a,b);
         }
         tarjan_init(n);
         suodian();
         int x=,y=;
         for(int i=;i<=taj;i++)
         {
             if(du[i]==) x++; //出度为0点的个数
             if(G[i].size()==) y++;
         }
         //printf("%d\n",j);
         if(taj==) printf("0\n");
         else
         printf("%d\n",max(x,y));
     }
     return ;
 }