标签：linker 部分 string 复杂 sign define int logs turn
题意：给定一棵树，然后给定每条边的权值，问你有多少个点对满足路径的权和小于等于m。
析：直接枚举是肯定不行的，会TLE，利用分治的思想，我们可以把树按重心分成几部分，那么答案就是所有子树的点对都经过重心的，对于所有的子树的重心也是这样，对于经过重心的，可以先求出每个点都重心的距离，再排序，利用单调性进行计算，这样的话会算重了，多算了在同一棵子树上的情况，这样再减去就好了。时间复杂度O(n*logn*logn)。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
vector<P> G[maxn];
int sz[maxn], f[maxn];
int root, num, ans;
int d[maxn];
bool vis[maxn];
vector<int> dist;
void dfs_for_root(int u, int fa){
sz[u] = 1; f[u] = 0;
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i].first;
if(v == fa || vis[v]) continue;
dfs_for_root(v, u);
sz[u] += sz[v];
f[u] = max(f[u], sz[v]);
}
f[u] = max(f[u], num - sz[u]);
if(f[u] < f[root]) root = u;
}
void dfs_for_dist(int u, int fa){
dist.push_back(d[u]);
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i].first;
if(v == fa || vis[v]) continue;
d[v] = d[u] + G[u][i].second;
dfs_for_dist(v, u);
}
}
int solve(int u, int val){
dist.clear();
int res = 0;
d[u] = val;
dfs_for_dist(u, -1);
sort(dist.begin(), dist.end());
for(int l = 0, r = (int)dist.size()-1; l < r; )
if(dist[l] + dist[r] <= m) res += r - l++;
else --r;
return res;
}
void dfs_for_ans(int u){
ans += solve(u, 0);
vis[u] = true;
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i].first;
if(vis[v]) continue;
ans -= solve(v, G[u][i].second);
root = 0;
f[0] = num = sz[v];
dfs_for_root(v, u);
dfs_for_ans(root);
}
}
int main(){
while(scanf("%d %d", &n, &m) == 2 && n+m){
for(int i = 1; i <= n; ++i) G[i].clear();
for(int i = 1; i < n; ++i){
int u, v, l;
scanf("%d %d %d", &u, &v, &l);
G[u].push_back(P(v, l));
G[v].push_back(P(u, l));
}
f[0] = num = n;
root = 0;
memset(vis, 0, sizeof vis);
dfs_for_root(1, root);
ans = 0;
dfs_for_ans(root);
printf("%d\n", ans);
}
return 0;
}
标签：linker 部分 string 复杂 sign define int logs turn
原文地址：http://www.cnblogs.com/dwtfukgv/p/7284775.html
