[hdu5416 CRB and Tree]树上路径异或和，dfs

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。

 

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#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;
}
posted @ 2015-08-21 11:16  jklongint  阅读(455)  评论(0)  编辑  收藏  举报