POJ 1987Distance Statistics

Description

Frustrated at the number of distance queries required to find a reasonable route for his cow marathon, FJ decides to ask queries from which he can learn more information. Specifically, he supplies an integer K (1 <= K <= 1,000,000,000) and wants to know how many pairs of farms lie at a distance at most K from each other (distance is measured in terms of the length of road required to travel from one farm to another). Please only count pairs of distinct farms (i.e. do not count pairs such as (farm #5, farm #5) in your answer). 

Input

* Lines 1 ..M+1: Same input format as in "Navigation Nightmare" 

* Line M+2: A single integer, K. 

Output

* Line 1: The number of pairs of farms that are at a distance of at most K from each-other. 

Sample Input

7 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S10

Sample Output

5

Hint

There are 5 roads with length smaller or equal than 10, namely 1-4 (3), 4-7 (2), 1-7 (5), 3-5 (7) and 3-6 (9). 

树分治，同POJ1741，这次又重新写了一遍，一A，感觉理解有深刻了一点。

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const double pi = acos(-1.0);const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int maxn = 1e5 + 10;int n, m, x, y, z, K;char ch[5];struct Tree{const static int maxn = 1e5 + 10;int ft[maxn], nt[maxn], u[maxn], v[maxn], sz, tot;int dis[maxn], vis[maxn], cnt[maxn], mx[maxn];void clear(int n){mx[sz = 0] = INF;for (int i = 1; i <= n; i++){ft[i] = -1;vis[i] = 0;}}void AddEdge(int x, int y, int z){u[sz] = y;v[sz] = z;nt[sz] = ft[x];ft[x] = sz++;u[sz] = x;v[sz] = z;nt[sz] = ft[y];ft[y] = sz++;}int dfs(int x, int fa, int sum){int z = mx[x] = 0;cnt[x] = 1;for (int i = ft[x]; i != -1; i = nt[i]){if (u[i] == fa || vis[u[i]]) continue;int y = dfs(u[i], x, sum);cnt[x] += cnt[u[i]];mx[x] = max(mx[x], cnt[u[i]]);if (mx[z] > mx[y]) z = y; }mx[x] = max(mx[x], sum - cnt[x]);return mx[x] < mx[z] ? x : z;}void dfsdis(int x, int fa, int len){dis[tot++] = len;for (int i = ft[x]; i != -1; i = nt[i]){if (u[i] == fa || vis[u[i]]) continue;dfsdis(u[i], x, len + v[i]);}}LL find(int x, int d){tot = 0;dfsdis(x, -1, d);sort(dis, dis + tot);LL ans = 0;for (int i = 0, j = tot - 1; i < j; i++){while (j > i&&dis[i] + dis[j] > K) j--;ans += j - i;}return ans;}LL work(int x, int sum){int y = dfs(x, -1, sum);LL ans = find(y, 0);vis[y] = 1;for (int i = ft[y]; i != -1; i = nt[i]){if (vis[u[i]]) continue;ans -= find(u[i], v[i]);ans += work(u[i], cnt[u[i]] > cnt[y] ? sum - cnt[y] : cnt[u[i]]);}return ans;}}solve;int main(){while (scanf("%d%d", &n, &m) != EOF){solve.clear(n);while (m--){scanf("%d%d%d%s", &x, &y, &z, ch);solve.AddEdge(x, y, z);}scanf("%d", &K);printf("%d\n", solve.work(1, n));}return 0;}