HDU 5039 Hilarity

题意：一棵树n个结点，每条边有0.1两种权值，每次询问权值为奇数的路径数目，或者改变某一条边的权值。

分析：这个题目很巧妙低利用了异或和的特性，dfs得到每个点到根结点的权值异或和，然后奇数则为1，偶数为0，异或和为0和1的点组成的路径权值和一定为奇数，询问结果就是0个数和1个数乘积2倍。

代码：

 #include <cstdio>
 #include <iostream>
 #include <vector>
 #include <map>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #define pb push_back
 #define mp make_pair

 #define lson   l, m, rt<<1
 #define rson   m+1, r, rt<<1|1
 #define sz(x) ((int)((x).size()))
 #define pb push_back
 #define in freopen("data.txt", "r", stdin);
 #define bug(x) printf("Line : %u >>>>>>\n", (x));
 #define inf 0x0f0f0f0f
 using namespace std;
 typedef pair<int, int> PII;
 typedef map<string, int> MPS;
 typedef long long LL;

 const int maxn =  + ;
 MPS mps;

 struct Edge {
     int u, v, c;
     Edge() {}
     Edge(int u, int v, int c):u(u), v(v), c(c) {}
 };
 vector<Edge> edges;
 vector<int> g[maxn];

 int n, m, q, cnt;
 int le[maxn], ri[maxn];

 int idx(char *s) {
     if(mps.count(s) == false)
         mps[s] = ++cnt;
     return mps[s];
 }

 void add(int u, int v, int c) {
     edges.pb(Edge(u, v, c));
     edges.pb(Edge(v, u, c));
     m = edges.size();
     g[u].pb(m-);
     g[v].pb(m-);
 }
 char sa[], sb[];
 int cover[maxn<<], val[maxn], dep[maxn], ans[maxn<<];
 void PushUp(int rt) {
     ans[rt] = ans[rt<<]+ans[rt<<|];
 }
 //线段树的每个结点要初始化！！！！

 void PushDown(int l, int r, int rt) {
     int m = (l+r)>>;
     if(cover[rt]) {
         cover[rt<<] ^= ;
         cover[rt<<|] ^= ;
         ans[rt<<] = m-l+-ans[rt<<];
         ans[rt<<|] = r-m-ans[rt<<|];
         cover[rt] = ;
     }
 }
 void build(int l, int r, int rt) {
     if(l == r) {
         cover[rt] = ;
         ans[rt] = val[l];
         return;
     }
     cover[rt] = ;
     int m = (l+r)>>;
     build(lson);
     build(rson);
     PushUp(rt);
 }
 void update(int l, int r, int rt, int L, int R) {
     if(L <= l && R >= r) {
         cover[rt] ^= ;
         ans[rt] = r-l+-ans[rt];
         return;
     }

     int m = (l+r)>>;
     PushDown(l, r, rt);
     if(L <= m)
         update(lson, L, R);
     if(R > m)
         update(rson, L, R);
     PushUp(rt);
 }
 void dfs(int u, int fa, int va) {
     dep[u] = dep[fa] + ;
     le[u] = ++cnt;
     val[cnt] = va;
     for(int i = ; i < (int)g[u].size(); i++) {
         Edge e = edges[g[u][i]];
         int v = e.v;
         int c = e.c;
         if(v == fa) continue;
         dfs(v, u, c^va);
     }
     ri[u] = cnt;
 }
 int main() {

     int T;
     int kase = ;
     for(int t = scanf("%d", &T); t <= T; t++) {
         scanf("%d", &n);
         mps.clear();
         cnt = ;
         edges.clear();
         for(int i = ; i <= n; i++) {
             g[i].clear();
             scanf("%s", sa);
             idx(sa);
         }
         for(int i = ; i < n-; i++) {
             int c, u, v;
             scanf("%s%s%d", sa, sb, &c);
             u = idx(sa);
             v = idx(sb);
             add(u, v, c);
         }
         scanf("%d", &q);
         printf("Case #%d:\n", ++kase);
         cnt = ;
         dfs(, , );
         build(, n, );
         LL res = ;
         while(q--) {
             scanf("%s", sa);
             if(sa[] == 'Q') {
                 res = (LL)(n-ans[])*ans[]*;
                 printf("%I64d\n", res);
             } else {
                 int x;
                 scanf("%d", &x);
                 x--;
                 x <<= ;
                 int u = edges[x].u;
                 int v = edges[x].v;
                 if(dep[u] < dep[v]) swap(u, v);
                 update(, n, , le[u], ri[u]);
             }
         }
     }
     return ;
 }

当然也有一种更复杂的方法，同Qtree4，可以拿来练手，感觉有点无聊。