BZOJ1954: Pku3764 The xor-longest Path

题解:

在树上i到j的异或和可以直接转化为i到根的异或和^j到根的异或和。

所以我们把每个点到根的异或和处理出来放到trie里面,再把每个点放进去跑一遍即可。

代码:

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 100000+5
 26 
 27 #define maxm 4000000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next)
 43 
 44 #define mod 1000000007
 45 
 46 using namespace std;
 47 
 48 inline int read()
 49 
 50 {
 51 
 52     int x=0,f=1;char ch=getchar();
 53 
 54     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 55 
 56     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
 57 
 58     return x*f;
 59 
 60 }
 61 struct edge{int go,next;ll w;}e[2*maxn];
 62 int n,tot,cnt,head[maxn],t[maxm][2];
 63 ll a[maxn];
 64 inline void insert(int x,int y,ll z)
 65 {
 66     e[++tot]=(edge){y,head[x],z};head[x]=tot;
 67     e[++tot]=(edge){x,head[y],z};head[y]=tot;
 68 }
 69 inline void dfs(int x,int fa)
 70 {
 71     for4(i,x)if((y=e[i].go)!=fa)
 72     {
 73         a[y]=a[x]^e[i].w;
 74         dfs(y,x);
 75     }
 76 }
 77 inline void add(ll x)
 78 {
 79     int now=1;
 80     for3(i,62,0)
 81     {
 82         int j=x>>i&1;
 83         if(!t[now][j])t[now][j]=++cnt;
 84         now=t[now][j];
 85     }
 86 }
 87 inline ll query(ll x)
 88 {
 89     int now=1;ll ret=0;
 90     for3(i,62,0)
 91     {
 92         int j=x>>i&1^1;
 93         if(t[now][j])ret|=(ll)1<<i;else j^=1;
 94         now=t[now][j];
 95     }
 96     return ret;
 97 }
 98 
 99 int main()
100 
101 {
102 
103     freopen("input.txt","r",stdin);
104 
105     freopen("output.txt","w",stdout);
106 
107     n=read();
108     for1(i,n-1){int x=read(),y=read(),z=read();insert(x,y,z);}
109     dfs(1,0);
110     cnt=1;
111     for1(i,n)add(a[i]);
112     ll ans=0;
113     for1(i,n)ans=max(ans,query(a[i]));
114     cout<<ans<<endl;
115 
116     return 0;
117 
118 }  
View Code

1954: Pku3764 The xor-longest Path

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 383  Solved: 161
[Submit][Status]

Description

给定一棵n个点的带权树,求树上最长的异或和路径

Input

The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.

Output

For each test case output the xor-length of the xor-longest path.

Sample Input

4
1 2 3
2 3 4
2 4 6

Sample Output

7

HINT

The xor-longest path is 1->2->3, which has length 7 (=3 ⊕ 4)
注意:结点下标从1开始到N....

Source

posted @ 2014-12-19 13:17  ZYF-ZYF  Views(240)  Comments(0Edit  收藏  举报