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 }

posted @ 2012-04-15 14:59 slon 阅读(633) 评论(6) 收藏举报

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