LCA UESTC 92 Journey

最新推荐文章于 2017-05-08 22:22:33 发布

weixin_30491641

最新推荐文章于 2017-05-08 22:22:33 发布

阅读量59

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Running-Time/p/4861585.html

本文探讨了在树形结构中加入额外边后，如何通过寻找最优路径来节省距离的问题。通过计算不同路径组合的最短路径，实现路径优化。

题目传送门

题意：先给一棵树，然后有一条额外的边，问u走到v从现在最短的路走和原来不加边走的路节省了多少距离

分析：首先跑不加边的树的LCA，这样能求出任意两点的距离，那么现在x和y多连了一条边，如果能节省路程那一定是走了xy这条边，那么暴力枚举组合，比如求u到v，新边xy，ans = min (ans, min (dux + dxy + dyv, duy + dyx + dxv))

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int D = 20;
const int INF = 0x3f3f3f3f;
struct Edge	{
	int v, w, nex;
	Edge (int v = 0, int w = 0, int nex = 0) : v (v), w (w), nex (nex) {}
}edge[N<<1];
int head[N], dep[N], rt[D][N];
int d[N];
int n, q, e;
int x, y, cost;

void init(void)	{
	memset (head, -1, sizeof (head));
	e = 0;
}

void add_edge(int u, int v, int w)	{
	edge[e] = Edge (v, w, head[u]);
	head[u] = e++;
}

void DFS(int u, int fa, int deep, int len)	{
	dep[u] = deep;	d[u] = len;	rt[0][u] = fa;
	for (int i=head[u]; ~i; i=edge[i].nex)	{
		int v = edge[i].v, w = edge[i].w;
		if (v == fa)	continue;
		DFS (v, u, deep + 1, len + w);
	}
}

int LCA(int u, int v)	{
	if (dep[u] < dep[v])	swap (u, v);
	for (int i=0; i<D; ++i)	{
		if ((dep[u] - dep[v]) >> i & 1)	{
			u = rt[i][u];
		}
	}
	if (u == v)	return u;
	for (int i=D-1; i>=0; --i)	{
		if (rt[i][u] != rt[i][v])	{
			u = rt[i][u];
			v = rt[i][v];
		}
	}
	return rt[0][u];
}

int solve(int u, int v)	{
	int lca = LCA (u, v);
	int ans = d[u] + d[v] - d[lca] * 2;

	int lca1 = LCA (u, x);
	int dux = d[u] + d[x] - d[lca1] * 2;

	int lca2 = LCA (u, y);
	int duy = d[u] + d[y] - d[lca2] * 2;

	int lca3 = LCA (v, x);
	int dvx = d[v] + d[x] - d[lca3] * 2;

	int lca4 = LCA (v, y);
	int dvy = d[v] + d[y] - d[lca4] * 2;

    int mn = min (dux + dvy + cost, duy + dvx + cost);
    if (mn > ans)	return 0;
    else	return ans - mn;
}

int main(void)	{
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		init ();
        scanf ("%d%d", &n, &q);
        for (int u, v, w, i=1; i<n; ++i)	{
			scanf ("%d%d%d", &u, &v, &w);
			add_edge (u, v, w);
			add_edge (v, u, w);
        }
        scanf ("%d%d%d", &x, &y, &cost);
        DFS (1, -1, 0, 0);
        for (int i=1; i<D; ++i)	{
			for (int j=1; j<=n; ++j)	{
				rt[i][j] = rt[i-1][j] == -1 ? -1 : rt[i-1][rt[i-1][j]];
			}
        }
        printf ("Case #%d:\n", ++cas);
        for (int u, v, i=1; i<=q; ++i)	{
			scanf ("%d%d", &u, &v);
			printf ("%d\n", solve (u, v));
        }
	}

	return 0;
}

转载于:https://www.cnblogs.com/Running-Time/p/4861585.html