UVALive 6900 Road Repair（点分治）

题意:

给定N≤2×104的一棵树,给定描述这棵树的四元组,(u,v,cost,profit),cost,profit≤103

对于树的所有简单路,找到一条简单路满足∑cost≤C≤2×107,且∑profit最大

求这个最大的∑profit

分析:

首先O(n2)暴力肯定要挂,dp背包的话显然C太大不行

我们需要一个带log的算法,考虑点分治

这题与POJ 1741 Tree很像,对于我们枚举的一条到重心的路径,假设这条路径的信息为(c,p)

此时我们需要知道当前子问题中其它到重心的路径(c′,p′)的max{p′},c′≤C−c,如何快速知道这个,可以考虑线段树或者BIT来维护最大值

那么ans=max{p+p′}

−−memset初始化BIT会T,正常回溯即可

代码:

//
//  Created by TaoSama on 2015-12-18
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 2e7 + 10;

int n, C, ans;

void getMax(int &x, int y) {x = max(x, y);}

struct BIT {
int mx[M];
void del(int i) {
for(; i < M; i += i & -i) mx[i] = 0;
}
void update(int i, int v) {
for(; i < M; i += i & -i) getMax(mx[i], v);
}
int query(int i) {
int ret = 0;
for(; i; i -= i & -i) getMax(ret, mx[i]);
return ret;
}
} bit;

struct Edge {
int v, nxt, c, b;
} edge[N << 1];
int head
, cnt;

void addEdge(int u, int v, int c, int b) {
edge[cnt] = (Edge) {v, head[u], c, b};
head[u] = cnt++;
edge[cnt] = (Edge) {u, head[v], c, b};
head[v] = cnt++;
}

bool vis
;
int sz
, mx
, centroid;

int getAll(int u, int f) {
int ret = 1;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(v == f || vis[v]) continue;
ret += getAll(v, u);
}
return ret;
}

void getCentroid(int u, int f, int all) {
sz[u] = 1; mx[u] = 0;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(v == f || vis[v]) continue;
getCentroid(v, u, all);
sz[u] += sz[v];
getMax(mx[u], sz[v]);
}
getMax(mx[u], all - sz[u]);
if(!centroid || mx[u] < mx[centroid]) centroid = u;
}

typedef pair<int, int> P;

void getPath(int u, int f, int cost, int profit, vector<P>& path) {
if(cost > C) return;
path.push_back(P(cost, profit));
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(v == f || vis[v]) continue;
getPath(v, u, cost + edge[i].c, profit + edge[i].b, path);
}
}

void solve(int u) {
centroid = 0;
getCentroid(u, -1, getAll(u, -1));
int s = centroid;
vis[s] = true;

for(int i = head[s]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;
solve(v);
}

vector<P> dummy;
for(int i = head[s]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;

vector<P> path;
getPath(v, s, edge[i].c, edge[i].b, path);
for(P &p : path) getMax(ans, p.second + bit.query(C - p.first));
for(P &p : path) bit.update(p.first, p.second);
dummy.insert(dummy.end(), path.begin(), path.end());
}

for(P &p : dummy) bit.del(p.first);
vis[s] = false;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
cnt = 0; memset(head, -1, sizeof head);
scanf("%d", &n);
for(int i = 1; i < n; ++i) {
int u, v, c, b; scanf("%d%d%d%d", &u, &v, &c, &b);
addEdge(u, v, c, b);
}
scanf("%d", &C);

ans = 0;
solve(1);
printf("%d\n", ans);
}
return 0;
}