POJ2114(树分治)

题意是给一棵最多1W个节点的树，求是否存在点对使得他们的路径是k。

类似于POJ的1741，修改下统计函数就可以了。注释掉的是一开始写的，很挫卡卡过，改掉以后效率好了一点。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>using namespace std;#define maxn 111111#define maxm 211111#define INF 11111111#define size Sizestruct node {    int from, to, next;    int w;}edge[maxm];int n, m, cnt, root;long long ans, k;int head[maxn];bool vis[maxn];int size[maxn], num[maxn]; //以i为根的子树的节点 i的子树中最多的节点long long d[maxn], Min;void add_edge (int from, int to, int w) {    edge[cnt].from = from, edge[cnt].to = to, edge[cnt].w = w, edge[cnt].next = head[from],    head[from] = cnt++;}void dfs_size (int u, int fa) {    size[u] = 1;    num[u] = 0;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (v == fa || vis[v])            continue;        dfs_size (v, u);        size[u] += size[v];        if (size[v] > num[u]) {            num[u] = size[v];        }    }}void find_root (int u, int fa, int pre) { //找到重心    long long cur = 0;    cur = max (num[u], size[pre]-size[u]);    if (cur < Min) {        Min = cur;        root = u;    }    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (v == fa || vis[v])            continue;        find_root (v, u, pre);    }}int tmp[maxn], tot;void get_dis (int u, int fa, int dis) { //得到每个点到当前根的距离    tmp[tot++] = dis;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (vis[v] || v == fa)            continue;        d[v] = dis+edge[i].w;        get_dis (v, u, d[v]);    }}int solve (int u, int d) {    long long ans = 0;    tot = 0;    get_dis (u, 0, d);    sort (tmp, tmp+tot);    int r = tot-1, l = 0;    while (r>l) {        for (; tmp[r]+tmp[l] > k && r > l; r--) {}        /*if (tmp[l]+tmp[r] == k) {            for (int i = r; i > l && tmp[l]+tmp[i] == k; i--)                ans++;        }*/        if (tmp[l]+tmp[r] == k) {            int cnt1 = 1, cnt2 = 1;            while (r > l && tmp[r-1] == tmp[r]) {                r--;                if (r > l) cnt2++;                else break;            }            while (r > l && tmp[l+1] == tmp[l]) {                l++;                if (r > l) cnt1++;                else break;            }            ans += cnt1*cnt2;        }        l++;    }    return ans;}void dfs (int u) {    Min = INF;    dfs_size (u, 0);    find_root (u, 0, u);    vis[root] = 1;    ans += solve (root, 0);    for (int i = head[root]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (vis[v])            continue;        ans -= solve (v, edge[i].w);        dfs (v);    }}int main () {    //freopen ("in", "r", stdin);    while (scanf ("%d", &n) == 1 && n) { //cout << n << endl;        cnt = 0;        int u, v;        long long w;        memset (head, -1, sizeof head);        for (u = 1; u <= n; u++) {            for (; ;) {                scanf ("%d", &v);                if (!v)                    break;                scanf ("%lld", &w); //cout << u << " " << v << " " <<w << endl;                add_edge (u, v, w);                add_edge (v, u, w);            }        }        for (; ;) {            scanf ("%lld\n", &k);            if (k == 0)                break;            memset (vis, 0, sizeof vis);            ans = 0;            dfs (1);            if (ans) printf ("AYE\n");            else printf ("NAY\n");        }        printf (".\n");        //cout << k << endl;    }    //printf (".\n");    return 0;}/*42 0 3 0 04 000120101230*/