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。