hdu 1269 迷宫城堡

迷宫城堡

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6826    Accepted Submission(s): 3034

Problem Description

为了训练小希的方向感，Gardon建立了一座大城堡，里面有N个房间(N<=10000)和M条通道(M<=100000)，每个通道都是单向的，就是说若称某通道连通了A房间和B房间，只说明可以通过这个通道由A房间到达B房间，但并不说明通过它可以由B房间到达A房间。Gardon需要请你写个程序确认一下是否任意两个房间都是相互连通的，即：对于任意的i和j，至少存在一条路径可以从房间i到房间j，也存在一条路径可以从房间j到房间i。

 

 

Input

输入包含多组数据，输入的第一行有两个数：N和M，接下来的M行每行有两个数a和b，表示了一条通道可以从A房间来到B房间。文件最后以两个0结束。

 

 

Output

对于输入的每组数据，如果任意两个房间都是相互连接的，输出"Yes"，否则输出"No"。

 

 

Sample Input

3 3 1 2 2 3 3 1 3 3 1 2 2 3 3 2 0 0

 

 

Sample Output

Yes

No

 

 

Author

Gardon

 

 

Source

HDU 2006-4 Programming Contest 

 

解题：强连通图判断，如果当前迷宫是一个强连通图，那么即Yes

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define INF 0x3f3f3f3f

15 using namespace std;

16 const int maxv = 10010;

17 const int maxe = 100010;

18 int dfn[maxv],low[maxv],n,m,index,scc;

19 vector<int>g[maxv];

20 bool instack[maxv];

21 stack<int>s;

22 void tarjan(int u){

23     s.push(u);

24     instack[u] = true;

25     dfn[u] = low[u] = ++index;

26     for(int i = 0; i < g[u].size(); i++){

27         int v = g[u][i];

28         if(!dfn[v]){

29             tarjan(v);

30             low[u] = min(low[u],low[v]);

31         }else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];

32     }

33     if(dfn[u] == low[u]){

34         int v;

35         scc++;

36         do{

37             v = s.top();

38             s.pop();

39         }while(v != u);

40     }

41 }

42 int main(){

43     int i,j,u,v;

44     while(scanf("%d%d",&n,&m),n+m){

45         for(i = 1; i <= n; i++)

46             g[i].clear();

47         memset(dfn,0,sizeof(dfn));

48         memset(low,0,sizeof(low));

49         memset(instack,false,sizeof(instack));

50         for(i = 0; i < m; i++){

51             scanf("%d%d",&u,&v);

52             g[u].push_back(v);

53         }

54         while(!s.empty()) s.pop();

55         index = scc = 0;

56         for(i = 1; i <= n; i++)

57             if(!dfn[i]) tarjan(i);

58         scc==1?puts("Yes"):puts("No");

59     }

60     return 0;

61 }

View Code

 

上面的代码有点傻逼，冗余了一些计算。。。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define INF 0x3f3f3f3f

15 using namespace std;

16 const int maxv = 10010;

17 const int maxe = 100010;

18 int dfn[maxv],low[maxv],n,m,index,scc;

19 vector<int>g[maxv];

20 bool instack[maxv],flag;

21 stack<int>s;

22 void tarjan(int u){

23     s.push(u);

24     instack[u] = true;

25     dfn[u] = low[u] = ++index;

26     for(int i = 0; i < g[u].size(); i++){

27         int v = g[u][i];

28         if(!dfn[v]){

29             tarjan(v);

30             low[u] = min(low[u],low[v]);

31         }else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];

32     }

33     if(dfn[u] == low[u]){

34         int v,o = 0;

35         do{

36             v = s.top();

37             s.pop();

38             o++;

39         }while(v != u);

40         if(o == n) flag = true;

41     }

42 }

43 int main(){

44     int i,j,u,v;

45     while(scanf("%d%d",&n,&m),n+m){

46         for(i = 1; i <= n; i++){

47             g[i].clear();

48             dfn[i] = 0;

49             low[i] = 0;

50             instack[i] = false;

51         }

52         for(i = 0; i < m; i++){

53             scanf("%d%d",&u,&v);

54             g[u].push_back(v);

55         }

56         while(!s.empty()) s.pop();

57         flag = index = 0;

58         tarjan(1);

59         flag?puts("Yes"):puts("No");

60     }

61     return 0;

62 }

View Code

 

美观点

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define INF 0x3f3f3f3f

15 using namespace std;

16 const int maxv = 10010;

17 const int maxe = 100010;

18 int dfn[maxv],low[maxv],n,m,index,scc;

19 vector<int>g[maxv];

20 bool instack[maxv];

21 stack<int>s;

22 bool tarjan(int u) {

23     s.push(u);

24     instack[u] = true;

25     dfn[u] = low[u] = ++index;

26     for(int i = 0; i < g[u].size(); i++) {

27         int v = g[u][i];

28         if(!dfn[v]) {

29             if(tarjan(v)) return true;

30             low[u] = min(low[u],low[v]);

31         } else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];

32     }

33     if(dfn[u] == low[u]) {

34         int v,o = 0;

35         do {

36             v = s.top();

37             s.pop();

38             o++;

39         } while(v != u);

40         if(o == n) return true;

41     }

42     return false;

43 }

44 int main() {

45     int i,j,u,v;

46     while(scanf("%d%d",&n,&m),n+m) {

47         for(i = 1; i <= n; i++) {

48             g[i].clear();

49             dfn[i] = 0;

50             low[i] = 0;

51             instack[i] = false;

52         }

53         for(i = 0; i < m; i++) {

54             scanf("%d%d",&u,&v);

55             g[u].push_back(v);

56         }

57         while(!s.empty()) s.pop();

58         index = 0;

59         tarjan(1)?puts("Yes"):puts("No");

60     }

61     return 0;

62 }

View Code