P4149 距离为K的点对(最少边数) n=200000 点分治

这题数据范围变成了200000 n^2就过不了 同时要求求的是最少的边数 不能容斥

#include
using namespace std;
const int MAXN = 2e5 + 5;
const int MAXM = 2e5 + 5;
int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], ed = 1;
int cost[MAXM << 1];
int ok[1000005];
inline void addedge(int u, int v, int c) {
        to[++ed] = v;
        cost[ed] = c;
        nxt[ed] = Head[u];
        Head[u] = ed;
}
inline void ADD(int u, int v, int c) {
        addedge(u, v, c);
        addedge(v, u, c);
}
int n, anser, k, cnt;
int sz[MAXN], f[MAXN], dep[MAXN], sumsz, root;
bool vis[MAXN];
pair<int, int> o[MAXN];
int num[MAXN];
void getroot(int x, int fa) {
        sz[x] = 1;
        f[x] = 0;
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (v == fa || vis[v]) {
                        continue;
                }
                getroot(v, x);
                sz[x] += sz[v];
                f[x] = max(f[x], sz[v]);
        }
        f[x] = max(f[x], sumsz - sz[x]);
        if (f[x] < f[root]) {
                root = x;
        }
}
void getdeep(int x, int fa, int deep) {
        if (dep[x] > k) {
                return;
        }
        o[++cnt] = make_pair(dep[x], deep);
        num[++num[0]] = dep[x];
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (v == fa || vis[v]) {
                        continue;
                }
                dep[v] = dep[x] + cost[i];
                getdeep(v, x, deep + 1);
        }
}
void calc(int x, int d) {
        num[0] = 0;
        dep[x] = d;
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (vis[v]) {
                        continue;
                }
                cnt = 0;
                dep[v] = dep[x] + cost[i];
                getdeep(v, x, 1);
                for (int j = 1; j <= cnt; j++) {
                        if (o[j].first <= k) {
                                if (ok[k - o[j].first] != INT_MAX) {
                                        anser = min(anser, ok[k - o[j].first] + o[j].second);
                                }
                        }
                }
                for (int j = 1; j <= cnt; j++) {
                        if (o[j].first <= k) {
                                ok[o[j].first] = min(o[j].second, ok[o[j].first]);
                        }
                }
        }
        for (int i = 1; i <= num[0]; i++) {
                ok[num[i]] = INT_MAX;
        }
}
void solve(int x) {
        vis[x] = 1;
        calc(x, 0);
        int totsz = sumsz;
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (vis[v]) {
                        continue;
                }
                root = 0;
                sumsz = sz[v] > sz[x] ? totsz - sz[x] : sz[v];
                getroot(v, 0);
                solve(root);
        }
}
int main() {
        scanf("%d %d", &n, &k);
        anser = INT_MAX;
        for (int i = 1; i <= k; i++) {
                ok[i] = INT_MAX;
        }
        cnt = 0;
        memset(Head, 0, sizeof(Head));
        memset(vis, 0, sizeof(vis));
        ed = 1;
        int u, v, c;
        for (int i = 1; i < n; i++) {
                scanf("%d %d %d", &u, &v, &c);
                ADD(u + 1, v + 1, c);
        }
        root = 0, sumsz = f[0] = n;
        getroot(1, 0);
        solve(root);
        printf("%d\n", anser == INT_MAX ? -1 : anser);
        return 0;
}

View Code

 注意dep[x]>k的时候要return 不然会re

转载于:https://www.cnblogs.com/Aragaki/p/10482758.html