hdu - 1269 迷宫城堡 (强连通裸题)
http://acm.hdu.edu.cn/showproblem.php?pid=1269
判断一个图是不是强连通,缩点之后判断顶点数是不是为1即可.
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <vector> 5 #include <cstring> 6 #include <algorithm> 7 #include <string> 8 #include <set> 9 #include <functional> 10 #include <numeric> 11 #include <sstream> 12 #include <stack> 13 #include <map> 14 #include <queue> 15 16 #define CL(arr, val) memset(arr, val, sizeof(arr)) 17 18 #define ll long long 19 #define inf 0x7f7f7f7f 20 #define lc l,m,rt<<1 21 #define rc m + 1,r,rt<<1|1 22 #define pi acos(-1.0) 23 24 #define L(x) (x) << 1 25 #define R(x) (x) << 1 | 1 26 #define MID(l, r) (l + r) >> 1 27 #define Min(x, y) (x) < (y) ? (x) : (y) 28 #define Max(x, y) (x) < (y) ? (y) : (x) 29 #define E(x) (1 << (x)) 30 #define iabs(x) (x) < 0 ? -(x) : (x) 31 #define OUT(x) printf("%I64d\n", x) 32 #define lowbit(x) (x)&(-x) 33 #define Read() freopen("a.txt", "r", stdin) 34 #define Write() freopen("dout.txt", "w", stdout); 35 36 using namespace std; 37 #define N 10100 38 //N为最大点数 39 #define M 100100 40 //M为最大边数 41 int n, m;//n m 为点数和边数 42 43 struct Edge{ 44 int from, to, nex; 45 bool sign;//是否为桥 46 }edge[M<<1]; 47 int head[N], edgenum; 48 void add(int u, int v){//边的起点和终点 49 Edge E={u, v, head[u], false}; 50 edge[edgenum] = E; 51 head[u] = edgenum++; 52 } 53 54 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) 55 int taj;//连通分支标号,从1开始 56 int Belong[N];//Belong[i] 表示i点属于的连通分支 57 bool Instack[N]; 58 vector<int> bcc[N]; //标号从1开始 59 60 void tarjan(int u ,int fa){ 61 DFN[u] = Low[u] = ++ Time ; 62 Stack[top ++ ] = u ; 63 Instack[u] = 1 ; 64 65 for (int i = head[u] ; ~i ; i = edge[i].nex ){ 66 int v = edge[i].to ; 67 if(DFN[v] == -1) 68 { 69 tarjan(v , u) ; 70 Low[u] = min(Low[u] ,Low[v]) ; 71 if(DFN[u] < Low[v]) 72 { 73 edge[i].sign = 1;//为割桥 74 } 75 } 76 else if(Instack[v]) Low[u] = min(Low[u] ,DFN[v]) ; 77 } 78 if(Low[u] == DFN[u]){ 79 int now; 80 taj ++ ; bcc[taj].clear(); 81 do{ 82 now = Stack[-- top] ; 83 Instack[now] = 0 ; 84 Belong [now] = taj ; 85 bcc[taj].push_back(now); 86 }while(now != u) ; 87 } 88 } 89 90 void tarjan_init(int all){ 91 memset(DFN, -1, sizeof(DFN)); 92 memset(Instack, 0, sizeof(Instack)); 93 top = Time = taj = 0; 94 for(int i=1;i<=all;i++)if(DFN[i]==-1 )tarjan(i, i); //注意开始点标!!! 95 } 96 vector<int>G[N]; 97 int du[N]; 98 void suodian(){ 99 memset(du, 0, sizeof(du)); 100 for(int i = 1; i <= taj; i++)G[i].clear(); 101 for(int i = 0; i < edgenum; i++){ 102 int u = Belong[edge[i].from], v = Belong[edge[i].to]; 103 if(u!=v) 104 { 105 G[u].push_back(v), du[v]++; 106 //printf("%d %d\n",u,v); 107 } 108 } 109 } 110 void init(){memset(head, -1, sizeof(head)); edgenum=0;} 111 112 int main() 113 { 114 // Read(); 115 int a,b; 116 while(~scanf("%d%d",&n,&m)&&n+m) 117 { 118 init(); 119 for(int i=0;i<m;i++) 120 { 121 scanf("%d%d",&a,&b); 122 add(a,b); 123 } 124 tarjan_init(n); 125 if(taj==1) puts("Yes"); 126 else puts("No"); 127 //suodian(); 128 } 129 return 0; 130 }