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数据范围是N:20w, K100w. 点分治, 我们只需考虑经过当前树根的方案. K最大只有100w, 直接开个数组CNT[x]表示与当前树根距离为x的最少边数, 然后就可以对根的子树依次dfs并更新CNT数组和答案.
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#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
const int maxn = 200009;
const int maxk = 1000009;
const int INF = 0x3F3F3F3F;
inline int read() {
int ret = 0;
char c = getchar();
for(; !isdigit(c); c = getchar());
for(; isdigit(c); c = getchar())
ret = ret * 10 + c - ‘0‘;
return ret;
}
int N, K, size[maxn], Rt, best, n, CNT[maxk], _n, ANS = INF;
pii T[maxn];
bool vis[maxn];
struct edge {
int to, w;
edge* next;
} E[maxn << 1], *pt = E, *head[maxn];
inline void add(int u, int v, int w) {
pt->to = v; pt->w = w; pt->next = head[u]; head[u] = pt++;
}
inline void addedge(int u, int v, int w) {
add(u, v, w); add(v, u, w);
}
void dfs(int x, int fa = -1) {
size[x] = 1;
int mx = 0;
for(edge* e = head[x]; e; e = e->next) if(e->to != fa && !vis[e->to]) {
dfs(e->to, x);
size[x] += size[e->to];
mx = max(mx, size[e->to]);
}
if((mx = max(mx, n - size[x])) < best) Rt = x, best = mx;
}
void DFS(int x, int dist, int cnt, int fa) {
if(dist > K) return;
ANS = min(ANS, cnt + CNT[K - dist]);
T[_n++] = make_pair(dist, cnt++);
for(edge* e = head[x]; e; e = e->next) if(e->to != fa && !vis[e->to])
DFS(e->to, dist + e->w, cnt, x);
}
void solve(int x) {
best = INF; dfs(x); x = Rt;
int p = _n = 0;
for(edge* e = head[x]; e; e = e->next) if(!vis[e->to]) {
DFS(e->to, e->w, 1, x);
for(int i = p; i < _n; i++)
CNT[T[i].first] = min(CNT[T[i].first], T[i].second);
p = _n;
}
for(int i = 0; i < _n; i++) CNT[T[i].first] = INF; CNT[0] = 0;
vis[x] = true;
for(edge* e = head[x]; e; e = e->next) if(!vis[e->to]) {
n = size[e->to];
solve(e->to);
}
}
void init() {
N = read(); K = read();
for(int i = 1; i < N; i++) {
int u = read(), v = read(), w = read();
addedge(u, v, w);
}
}
int main() {
init();
n = N;
memset(CNT, INF, sizeof CNT); CNT[0] = 0;
solve(0);
printf("%d\n", ANS != INF ? ANS : -1);
return 0;
}
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2599: [IOI2011]Race
Time Limit: 50 Sec Memory Limit: 128 MB
Submit: 1806 Solved: 538
[Submit][Status][Discuss]
Description
给一棵树,每条边有权.求一条路径,权值和等于K,且边的数量最小.
Input
第一行 两个整数 n, k
第二..n行 每行三个整数 表示一条无向边的两端和权值 (注意点的编号从0开始)
Output
一个整数 表示最小边数量 如果不存在这样的路径 输出-1
Sample Input
4 3
0 1 1
1 2 2
1 3 4
Sample Output
2
HINT
Source
BZOJ 2599: [IOI2011]Race( 点分治 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/4803820.html
