标签：ever most uil can ble code res rebuild div
InputOn the first line one positive number: the number of testcases, at most 100. After that per testcase:
* One line containing two integers n (1 ≤ n ≤ 20000) and m (0 ≤ m ≤ 50000): the number of statements and the number of implications that have already been proved.
* m lines with two integers s1 and s2 (1 ≤ s1, s2 ≤ n and s1 ≠ s2) each, indicating that it has been proved that statement s1 implies statement s2.OutputPer testcase:
* One line with the minimum number of additional implications that need to be proved in order to prove that all statements are equivalent.Sample Input
2 4 0 3 2 1 2 1 3
Sample Output
4 2
题意：
给定一张有向图，问最少添加几条边使得有向图成为一个强连通图。
题解：
缩完点的图是一个DAG，变成强联通就是，一个点至少一个出度一个入度
所以只需要输出缩完点后的图入度和出度最大值既可。
这个真的很好想，自己瞎比比搞了半天，浪费了许多时间。
真的菜。
想到后怒删代码，修改就过了。
1 #include<cstring> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstdio> 6 #define N 20007 7 #define M 50007 8 using namespace std; 9 10 int n,m,tim,sc,totalin,totalout; 11 int top,dfn[N],low[N],stack[N],ins[N],bel[N],chu[N],ru[N],boo[N]; 12 int cnt,head[N],Next[M],rea[M]; 13 struct Node 14 { 15 int ru,chu; 16 void init() 17 { 18 ru=chu=0; 19 } 20 }zhi[N]; 21 22 void add(int u,int v) 23 { 24 Next[++cnt]=head[u]; 25 head[u]=cnt; 26 rea[cnt]=v; 27 } 28 void Tarjan(int u) 29 { 30 dfn[u]=low[u]=++tim; 31 stack[++top]=u,ins[u]=true; 32 for (int i=head[u];i!=-1;i=Next[i]) 33 { 34 int v=rea[i]; 35 if (!dfn[v]) 36 { 37 Tarjan(v); 38 low[u]=min(low[u],low[v]); 39 } 40 else if (ins[v]) low[u]=min(low[u],dfn[v]); 41 } 42 if (low[u]==dfn[u]) 43 { 44 sc++;int x=-1; 45 while(x!=u) 46 { 47 x=stack[top--]; 48 ins[x]=0; 49 bel[x]=sc; 50 } 51 } 52 } 53 void rebuild() 54 { 55 for (int u=1;u<=n;u++) 56 { 57 for (int i=head[u];i!=-1;i=Next[i]) 58 { 59 int v=rea[i]; 60 if (bel[v]!=bel[u]) 61 { 62 chu[bel[u]]++; 63 ru[bel[v]]++; 64 } 65 } 66 } 67 for (int i=1;i<=sc;i++) 68 { 69 if (!chu[i]) totalout++; 70 if (!ru[i]) totalin++; 71 } 72 } 73 int main() 74 { 75 int T;scanf("%d",&T); 76 while (T--) 77 { 78 cnt=sc=0,top=0,totalin=totalout=0; 79 memset(head,-1,sizeof(head)); 80 memset(dfn,0,sizeof(dfn)); 81 memset(low,0,sizeof(low)); 82 memset(boo,0,sizeof(boo)); 83 memset(chu,0,sizeof(chu)); 84 memset(ru,0,sizeof(ru)); 85 scanf("%d%d",&n,&m); 86 for (int i=1,x,y;i<=m;i++) 87 { 88 scanf("%d%d",&x,&y); 89 add(x,y); 90 } 91 for (int i=1;i<=n;i++) 92 if (!dfn[i]) Tarjan(i); 93 rebuild(); 94 int ans=max(totalout,totalin); 95 if (ans==1) ans=0; 96 printf("%d\n",ans); 97 } 98 }
HDU 2767 Proving Equivalences(强连通 Tarjan+缩点)
标签：ever most uil can ble code res rebuild div
原文地址：http://www.cnblogs.com/fengzhiyuan/p/7747628.html
