zoj 3602 Count the Trees

 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3602

省赛的C题，方法是hash每个节点定义的状态。

关键貌似要直接DFS递归会爆栈，所以我手写了一个栈模拟。

下面还是贴没有模拟栈之前的代码，提交会sf，不过看起来会好理解很多，模拟部分可以自己去写。

View Code

 1 #include<iostream> 

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<map>

 5 #include<cstring>

 6     using namespace std;

 7 typedef long long LL;

 8 #define N 100010

 9 struct node{

10     int l,r;

11     friend bool operator<(const node &x,const node &y){

12         if(y.l!=x.l) return y.l<x.l;

13         return y.r<x.r;

14     }

15 }tree[2][N];

16 int n,m,cs,num[2][N];

17 map<node,int> state;

18 int id(node x){

19     if(state.find(x)==state.end())

20         state[x]=++cs;

21     return state[x];

22 }

23 int dfs(int u,int kind){

24     if(u==-1) return 0;

25     node temp;

26     temp.l=dfs(tree[kind][u].l,kind);

27     temp.r=dfs(tree[kind][u].r,kind);

28     int ans=id(temp);

29     num[kind][ans]++;

30     return ans;

31 }

32 int main(){

33     int T,a,b;

34     scanf("%d",&T);

35     while(T--){

36         scanf("%d%d",&n,&m);

37         for(int i=1;i<=n;++i){

38             scanf("%d%d",&a,&b);

39             tree[0][i].l=a;

40             tree[0][i].r=b;

41         }

42         for(int i=1;i<=m;++i){

43             scanf("%d%d",&a,&b);

44             tree[1][i].l=a;

45             tree[1][i].r=b;

46         }

47         state.clear();

48         cs=0;

49         memset(num,0,sizeof(num));

50         dfs(1,0);

51         dfs(1,1);

52         LL ans=0;

53         for(int i=0;i<=n+m;++i)

54             ans+=LL(num[0][i])*LL(num[1][i]);

55         printf("%lld\n",ans);

56     }

57     return 0;

58 }

 

这是模拟栈后的代码，感觉自己写的不好...

View Code

 1 #include<iostream> 

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<map>

 5 #include<cstring>

 6 using namespace std;

 7 typedef long long LL;

 8 #define N 100010

 9 struct node{

10     int l,r;

11     friend bool operator<(const node &x,const node &y){

12         if(y.l!=x.l) return y.l<x.l;

13         return y.r<x.r;

14     }

15 }tree[2][N];

16 int n,m,cs,num[2][N<<1];

17 map<node,int> state;

18 int id(node x){

19     if(state.find(x)==state.end())

20         state[x]=++cs;

21     return state[x];

22 }

23 struct stk{

24     int u,op,st;

25 }qu[N];

26 void dfs(int u,int kind){

27     int top=0;

28     qu[top].u=u;

29     qu[top++].op=0;

30     while(top){

31         stk now=qu[top-1];

32         if(now.u==-1){

33             qu[top-1].st=0;

34             top--;

35         }

36         else if(!now.op){

37             qu[top].u=tree[kind][now.u].l;

38             qu[top].op=0;

39             qu[top-1].op=1;

40             top++;

41         }

42         else if(now.op==1){

43             qu[top].u=tree[kind][now.u].r;

44             qu[top].op=0;

45             qu[top-1].op=2;

46             qu[top-1].st=qu[top].st;

47             top++;

48         }

49         else{

50             node temp;

51             temp.l=qu[top-1].st;

52             temp.r=qu[top].st;

53             int rig=id(temp);

54             qu[top-1].st=rig;

55             num[kind][rig]++;

56             top--;

57         }

58     }

59 }

60 int main(){

61     int T,a,b;

62     scanf("%d",&T);

63     while(T--){

64         scanf("%d%d",&n,&m);

65         for(int i=1;i<=n;++i){

66             scanf("%d%d",&a,&b);

67             tree[0][i].l=a;

68             tree[0][i].r=b;

69         }

70         for(int i=1;i<=m;++i){

71             scanf("%d%d",&a,&b);

72             tree[1][i].l=a;

73             tree[1][i].r=b;

74         }

75         state.clear();

76         cs=0;

77         memset(num,0,sizeof(num));

78         dfs(1,0);

79         dfs(1,1);

80         LL ans=0;

81         for(int i=0;i<=cs;++i)

82             ans+=LL(num[0][i])*LL(num[1][i]);

83         printf("%lld\n",ans);

84     }

85     return 0;

86 }