Ural 1557 Network Attack
题意是在一个联通的无向图中切断两条边使其不连通,求切边的放法数。。。
具体参加曹钦翔神犇的神论文吧 http://www.docin.com/p-52577984.html
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<string> 5 #include<iostream> 6 #include<vector> 7 using namespace std; 8 const int maxn = 2005; 9 const int maxm = 200005; 10 int nume,ct,n,m; 11 int g[maxn]; 12 int dfn[maxn]; 13 int s[maxn],t[maxn]; 14 int ma[maxn][maxn]; 15 long long ans; 16 vector<int>a[maxn]; 17 int tot; 18 struct edge 19 { 20 int v,nxt; 21 }; 22 edge e[maxm]; 23 void init() 24 { 25 nume = 1; 26 memset(g,0,sizeof(g)); 27 } 28 void addedge(int u,int v) 29 { 30 nume++; 31 e[nume].v = v; 32 e[nume].nxt = g[u]; 33 g[u] = nume; 34 } 35 void dfs(int x,int edg) 36 { 37 ct++; 38 dfn[x] = ct; 39 int sum1 =0,sum2 = 0; 40 for (int i = g[x];i; i = e[i].nxt) 41 { 42 if (i == edg || edg == (i ^ 1)) continue; 43 int y = e[i].v; 44 if (!dfn[y]) 45 { 46 a[x].push_back(y); 47 sum1++; 48 dfs(y,i); 49 s[x] = s[x] + s[y]; 50 //cout<<x<<" "<<y<<" "<<s[x]<<" "<<s[y]<<endl; 51 for (int j = 1; j < dfn[x]; j++) ma[x][j] += ma[y][j]; 52 } 53 else 54 { 55 if (dfn[y] < dfn[x]) 56 { 57 sum2++; 58 ma[x][dfn[y]]++; 59 } 60 else sum1++; 61 } 62 } 63 //cout<<x<<" "<<sum1<<" "<<sum2<<endl; 64 s[x] = s[x] - sum1 + sum2; 65 } 66 void check(int x,int y) 67 { 68 for (int i = 0; i < a[x].size(); i++) 69 { 70 if (s[y] == s[a[x][i]] && t[a[x][i]] < dfn[y]) ans++; 71 check(a[x][i],y); 72 } 73 } 74 int main() 75 { 76 while (scanf("%d%d",&n,&m) != EOF) 77 { 78 init(); 79 for (int i = 1; i <= n; i++) a[i].clear(); 80 int x,y; 81 for (int i = 1; i <= m; i++) 82 { 83 scanf("%d%d",&x,&y); 84 if (x == y) continue; 85 addedge(x,y); 86 addedge(y,x); 87 } 88 memset(dfn,0,sizeof(dfn)); 89 memset(s,0,sizeof(s)); 90 memset(t,0,sizeof(t)); 91 memset(ma,0,sizeof(ma)); 92 ct = 0; 93 for (int i = 2;i <= n; i++) s[i] = 1; 94 dfs(1,-1); 95 //for (int i = 1; i <= n; i++) cout<<i<<" "<<s[i]<<endl; 96 for (int i = 2; i <= n; i++) 97 for (int j = dfn[i] - 1; j >= 1; j--) 98 if (ma[i][j] > 0) 99 { 100 t[i] = j; 101 break; 102 } 103 ans = 0; 104 tot = 0; 105 for (int i = 2; i <= n; i++) 106 { 107 if (s[i] == 1) tot++; 108 else 109 { 110 if (s[i] == 2) ans++; 111 check(i,i); 112 } 113 } 114 long long tmp = tot; 115 tmp = tmp * (m - tmp) + tmp * (tmp - 1) / 2; 116 ans = ans + tmp; 117 printf("%I64d\n",ans); 118 } 119 return 0; 120 }
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