| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 3533 | Accepted: 988 |
Description
After recapturing Sylla, the Company plans to establish a new secure system, a transferring net! The new system is designed as follows:
The Company staff choose N cities around the nation which are connected by "security tunnels" directly or indirectly. Once a week, Sylla is to be transferred to another city through the tunnels. As General ordered, the transferring net must reach a certain security level that there are at least 3 independent paths between any pair of cities a, b. When General says the paths are independent, he means that the paths share only a and b in common.
Given a design of a transferring net, your work is to inspect whether it reaches such security level.
Input
The input consists of several test cases.
For each test case, the first line contains two integers, N ≤ 500 and M ≤ 20000. indicating the number of cities and tunnels.
The following M lines each contains two integers a and b (0 ≤ a, b < N), indicating the city a and city b are connected directly by a tunnel.
The input ends by two zeroes.
Output
For each test case output "YES" if it reaches such security level, "NO" otherwise.
Sample Input
4 6 0 1 0 2 0 3 1 2 1 3 2 3 4 5 0 1 0 2 0 3 1 2 1 3 7 6 0 1 0 2 0 3 1 2 1 3 2 3 0 0
Sample Output
YES NO NO
思路：判断给定图是否为三联通分量；
以下情况不是：
1.图不连通
2.图有割点
3.删去一个点后存在割点。
这题我用了读入挂（抄的别人的板子）才过。。。。
代码：
1 #include<iostream> 2 #include<algorithm> 3 #include<vector> 4 #include<stack> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<cstdio> 9 #include<cstring> 10 #include<cmath> 11 #include<ctime> 12 #define fuck(x) cout<<#x<<" = "<<x<<endl; 13 #define ls (t<<1) 14 #define rs ((t<<1)+1) 15 using namespace std; 16 typedef long long ll; 17 typedef unsigned long long ull; 18 const int maxn = 50086; 19 const int inf = 2.1e9; 20 const ll Inf = 999999999999999999; 21 const int mod = 1000000007; 22 const double eps = 1e-6; 23 const double pi = acos(-1); 24 int f[1000]; 25 int index,root,n,m,DEL; 26 int cnt,Head[1000],v[maxn],Next[maxn]; 27 void init() 28 { 29 cnt=0;root=1; 30 memset(Head,-1,sizeof(Head)); 31 for(int i=0;i<=n;i++){ 32 f[i]=i; 33 } 34 } 35 36 inline bool scan_d(int &num) 37 { 38 char in;bool IsN=false; 39 in=getchar(); 40 if(in==EOF) return false; 41 while(in!=‘-‘&&(in<‘0‘||in>‘9‘)) in=getchar(); 42 if(in==‘-‘){ IsN=true;num=0;} 43 else num=in-‘0‘; 44 while(in=getchar(),in>=‘0‘&&in<=‘9‘){ 45 num*=10,num+=in-‘0‘; 46 } 47 if(IsN) num=-num; 48 return true; 49 } 50 void add(int x,int y) 51 { 52 cnt++; 53 v[cnt]=y; 54 Next[cnt]=Head[x]; 55 Head[x]=cnt; 56 57 cnt++; 58 v[cnt]=x; 59 Next[cnt]=Head[y]; 60 Head[y]=cnt; 61 } 62 63 int num[508],low[508]; 64 bool dfs(int cur,int father) 65 { 66 int child=0; 67 index++; 68 num[cur]=index; 69 low[cur]=index; 70 for(int k=Head[cur];k!=-1;k=Next[k]){ 71 if(v[k]==DEL){continue;} 72 // fuck(v[k]) 73 if(num[v[k]]==0){ 74 child++; 75 if(!dfs(v[k],cur)){return false;} 76 low[cur]=min(low[cur],low[v[k]]); 77 if(cur!=root&&low[v[k]]>=num[cur]){ 78 return false; 79 } 80 if(cur==root&&child==2){ 81 return false; 82 } 83 } 84 else if(v[k]!=father){ 85 low[cur]=min(low[cur],num[v[k]]); 86 } 87 } 88 return true; 89 } 90 91 bool solve() 92 { 93 memset(num,0,sizeof(num)); 94 memset(low,0,sizeof(low)); 95 root=2; 96 DEL=1; 97 if(!dfs(2,root)){return false;} 98 root=1; 99 for(int i=2;i<=n;i++){ 100 memset(num,0,sizeof(num)); 101 memset(low,0,sizeof(low)); 102 DEL=i; 103 if(!dfs(1,root)){ 104 // cout<<i<<endl; 105 return false; 106 } 107 } 108 return true; 109 } 110 111 int getf(int x) 112 { 113 if(f[x]==x){return x;} 114 return f[x]=getf(f[x]) 115 ;} 116 117 bool Merge(int x,int y) 118 { 119 int s1=getf(x); 120 int s2=getf(y); 121 if(s1!=s2){ 122 f[s1]=s2; 123 return true; 124 } 125 return false; 126 } 127 128 int main() 129 { 130 // ios::sync_with_stdio(false); 131 // freopen("in.txt","r",stdin); 132 133 while(scanf("%d%d",&n,&m)!=EOF&&n&&m){ 134 init(); 135 int snum = 0; 136 for(int i=1;i<=m;i++){ 137 int x,y; 138 scan_d(x);scan_d(y); 139 add(x+1,y+1); 140 if(Merge(x+1,y+1)){ 141 snum++; 142 } 143 } 144 if(snum!=n-1){printf("NO\n");continue;} 145 memset(num,0,sizeof(num)); 146 memset(low,0,sizeof(low)); 147 root=1; 148 DEL=0; 149 if(!dfs(1,root)){printf("NO\n");continue;} 150 151 152 if(solve()){printf("YES\n");} 153 else printf("NO\n"); 154 155 } 156 157 return 0; 158 }