您的位置:首页 > 其它

[hdu5416 CRB and Tree]树上路径异或和,dfs

2015-08-21 11:16 344 查看
题意:给一棵树,每条边有一个权值,求满足u到v的路径上的异或和为s的(u,v)点对数

思路:计a到b的异或和为f(a,b),则f(a,b)=f(a,root)^f(b,root)。考虑dfs,一边计算当前点到根的f值,用一个数组记录当前遍历过的点中到根的异或值为i的点的个数,那么答案可以O(1)算出来,更新也是O(1)的。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129

#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

const int maxn = 2e5 + 7;

struct Graph {
vector<vector<int> > G;
void clear() { G.clear(); }
void resize(int n) { G.resize(n + 2); }
void add(int u, int v) { G[u].push_back(v); }
vector<int> & operator [] (int u) { return G[u]; }
};
Graph G;

struct Edge {
int u, v, w;
Edge(int u, int v, int w) {
this->u = u;
this->v = v;
this->w = w;
}
};
vector<Edge> E;

bool vis[maxn];
int cnt[maxn];
int Q[20];
ll ans[20];
int q, now;

void add(int u, int v, int w) {
E.pb(Edge(u, v, w));
G.add(u, E.size() - 1);
}

void dfs(int u) {
cnt[now] ++;
for (int i = 0; i < q; i ++) {
ans[i] += cnt[now ^ Q[i]];
}
vis[u] = true;
for (int i = 0; i < G[u].size(); i ++) {
Edge e = E[G[u][i]];
if (!vis[e.v]) {
now ^= e.w;
dfs(e.v);
now ^= e.w;
}
}
}

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int T, n;
cin >> T;
while (T --) {
cin >> n;
E.clear();
G.clear();
G.resize(n);
for (int i = 1; i < n; i ++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
cin >> q;
for (int i = 0; i < q; i ++) {
scanf("%d", Q + i);
}
fillchar(vis, 0);
now = 0;
fillchar(cnt, 0);
fillchar(ans, 0);
dfs(1);
for (int i = 0; i < q; i ++) {
cout << ans[i] << endl;
}
}
return 0;
}

内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: