题意:
与POJ-1741一致,传送门:
点击打开链接
思路:
没什么意思,直接套模板。
代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 10;
struct node {
int v;
ll w;
};
ll k;
int n, m, root, Max, ans;
vector <node> tree[MAXN];
vector <ll> dis;
int sz[MAXN], maxv[MAXN];
bool vis[MAXN];
void init() {
memset(vis, false, sizeof(vis));
for (int i = 1; i <= n; i++) tree[i].clear();
}
void dfs_size(int u, int pre) { // 求出每个子树的大小,以及每个节点的最大儿子
sz[u] = 1; maxv[u] = 0;
int cnt = tree[u].size();
for (int i = 0; i < cnt; i++) {
int v = tree[u][i].v;
if (v == pre || vis[v]) continue;
dfs_size(v, u);
sz[u] += sz[v];
maxv[u] = max(maxv[u], sz[v]);
}
}
void dfs_root(int r, int u, int pre) { // 找出以u为根的子树的重心
maxv[u] = max(maxv[u], sz[r] - sz[u]);
if (Max > maxv[u]) {
Max = maxv[u];
root = u;
}
int cnt = tree[u].size();
for (int i = 0; i < cnt; i++) {
int v = tree[u][i].v;
if (v == pre || vis[v]) continue;
dfs_root(r, v, u);
}
}
void dfs_dis(int u, int pre, int d) { // 求出当前子树中所有点到根的距离
dis.push_back(d);
int cnt = tree[u].size();
for (int i = 0; i < cnt; i++) {
int v = tree[u][i].v, w = tree[u][i].w;
if (v == pre || vis[v]) continue;
dfs_dis(v, u, d + w);
}
}
int cal(int u, int d) { // 计算当前子树中合法的点对数
int res = 0;
dis.clear(); dfs_dis(u, -1, d);
sort (dis.begin(), dis.end());
int i = 0, j = dis.size() - 1;
while (i < j) {
while (dis[i] + dis[j] > k && i < j) --j;
res += j - i;
++i;
}
return res;
}
void dfs(int u) { // 总的dfs求解
Max = n;
dfs_size(u, -1); dfs_root(u, u, -1);
ans += cal(root, 0);
vis[root] = true;
int cnt = tree[root].size(), rt = root; // 一定要注意这样里的root是全局变量,在递归之后可能改变,需要提前保存下来。
for (int i = 0; i < cnt; i++) {
int v = tree[rt][i].v, w = tree[rt][i].w;
if (vis[v]) continue;
ans -= cal(v, w);
dfs(v);
}
}
int main() {
//freopen("in", "r", stdin);
while (scanf("%d%d", &n, &m) == 2) {
init();
char s[10];
for (int i = 1; i < n; i++) {
int u, v;
ll w;
scanf("%d%d%lld%s", &u, &v, &w, s);
tree[u].push_back((node){v, w});
tree[v].push_back((node){u, w});
}
scanf("%lld", &k);
ans = 0;
dfs(1);
printf("%d\n", ans);
}
return 0;
}