标签：int scanf 路径 opera accept follow name cep body
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 6361 | Accepted: 1378 |
Description
In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p:
⊕ is the xor operator.
We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path?
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
Sample Input
4 0 1 3 1 2 4 1 3 6
Sample Output
7
Hint
The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)
Source
trie树+贪心
DFS预处理出每个结点到根的路径的异或和。两点之间路径的异或和等于各自到根的路径的异或和的异或。
将所有的异或和转化成二进制串，建成trie树。
对于每个二进制串，在trie树上贪心选取使异或值最大的路径（尽量通往数值相反的结点），记录最优答案。
↑为了防止匹配到自身的一部分，每个二进制串都应该先查完再插入trie树。
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<vector> 8 using namespace std; 9 const int mxn=200010; 10 int read(){ 11 int x=0,f=1;char ch=getchar(); 12 while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 13 while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 14 return x*f; 15 } 16 struct edge{ 17 int v,nxt,w; 18 }e[mxn<<1]; 19 int hd[mxn],mct=0; 20 void add_edge(int u,int v,int w){ 21 e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return; 22 } 23 int t[mxn*10][2],cnt=1; 24 int n,ans=0; 25 int f[mxn]; 26 void insert(int x){ 27 int now=1; 28 for(int i=31;i>=0;i--){ 29 int v=(x>>i)&1; 30 if(!t[now][v])t[now][v]=++cnt; 31 now=t[now][v]; 32 } 33 return; 34 } 35 void query(int x){ 36 int now=1; 37 int res=0; 38 for(int i=31;i>=0;i--){ 39 int v=(x>>i)&1; 40 if(t[now][v^1]){ 41 v^=1; 42 res+=(1<<i); 43 } 44 now=t[now][v]; 45 } 46 ans=max(res,ans); 47 return; 48 } 49 void DFS(int u,int fa,int num){ 50 f[u]=num; 51 for(int i=hd[u];i;i=e[i].nxt){ 52 int v=e[i].v; 53 if(v==fa)continue; 54 DFS(v,u,num^e[i].w); 55 } 56 return; 57 } 58 void init(){ 59 memset(hd,0,sizeof hd); 60 memset(t,0,sizeof t); 61 // memset(f,0,sizeof f); 62 mct=0;cnt=1;ans=0; 63 return; 64 } 65 int main(){ 66 while(scanf("%d",&n)!=EOF){ 67 init(); 68 int i,j,u,v,w; 69 for(i=1;i<n;i++){ 70 u=read();v=read();w=read(); 71 u++;v++; 72 add_edge(u,v,w); 73 add_edge(v,u,w); 74 } 75 DFS(1,0,0); 76 for(i=1;i<=n;i++){ 77 // printf("f[%d]:%d\n",i,f[i]); 78 query(f[i]); 79 insert(f[i]); 80 } 81 printf("%d\n",ans); 82 } 83 return 0; 84 }
标签：int scanf 路径 opera accept follow name cep body
原文地址：http://www.cnblogs.com/SilverNebula/p/6185373.html
