【BZOJ4326】NOIP2015 运输计划【LCA】【路径求并】

【题目链接】

大概并没有用正解，卡常数A了。

先二分答案mid，那么边权变为0的边一定在所有长度大于mid的路径的交上，且这条边的边权至少为路径长度减去mid（否则不可能长度不可能降低到mid之下）。

每次check，把长度大于mid的路径加进去，求一次交（一次dfs就求出来了），遍历每条边，看是否有边满足上述条件。

/* Pigonometry */
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 300005, maxm = 300005, maxk = 20;

int n, m, head[maxn], cnt;

struct _edge {
	int v, w, next;
} g[maxn << 1];

struct _line {
	int u, v, lca, dis;
} line[maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v, int w) {
	g[cnt] = (_edge){v, w, head[u]};
	head[u] = cnt++;
}

/* doubling lca */
int pre[maxn][maxk], val[maxn], depth[maxn], dis[maxn];

inline void dfs(int x) {
	for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ pre[x][0]) {
		pre[g[i].v][0] = x;
		val[g[i].v] = g[i].w;
		depth[g[i].v] = depth[x] + 1;
		dis[g[i].v] = dis[x] + g[i].w;
		dfs(g[i].v);
	}
}

inline int getlca(int u, int v) {
	if(depth[u] < depth[v]) swap(u, v);
	for(int i = maxk - 1; i >= 0; i--) if(depth[pre[u][i]] >= depth[v]) u = pre[u][i];
	for(int i = maxk - 1; i >= 0; i--) if(pre[u][i] != pre[v][i]) u = pre[u][i], v = pre[v][i];
	return u == v ? u : pre[u][0];
}

int sum[maxn];

inline void work(int x) {
	for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ pre[x][0]) {
		work(g[i].v);
		sum[x] += sum[g[i].v];
	}
}

inline bool check(int x) {
	int tot = 0, w = 0;
	for(int i = 1; i <= n; i++) sum[i] = 0;
	for(int i = 1; i <= m; i++) if(line[i].dis > x) {
		tot++;
		sum[line[i].u]++;
		sum[line[i].v]++;
		sum[line[i].lca] -= 2;
		w = max(w, line[i].dis - x);
	}
	work(1);
	for(int i = 1; i <= n; i++)
		if(sum[i] == tot && val[i] >= w) return 1;
	return 0;
}

int main() {
	n = iread(); m = iread();
	for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0;

	for(int i = 1; i < n; i++) {
		int u = iread(), v = iread(), w = iread();
		add(u, v, w); add(v, u, w);
	}

	dfs(1);
	for(int j = 1; j < maxk; j++) for(int i = 1; i <= n; i++)
		pre[i][j] = pre[pre[i][j - 1]][j - 1];

	int l = 0, r = 0;
	for(int i = 1; i <= m; i++) {
		int u = iread(), v = iread(), lca = getlca(u, v), d = dis[u] + dis[v] - 2 * dis[lca];
		line[i] = (_line){u, v, lca, d};
		r = max(r, d);
	}

	while(l <= r) {
		int mid = l + r >> 1;
		if(check(mid)) r = mid - 1;
		else l = mid + 1;
	}
	printf("%d\n", l);
	return 0;
}