HDU 2586 How far away(LCA+邻接表)

How far away

&题解：

和上篇是一样的题,这用的是lca方法做的, 不知道为什么,把数组开到80000 就a了 >_<
哈 我现在知道为什么了,因为我的rmq数组没有乘2,rmq比较的是depth数组,但是depth数组的范围是maxn*2,所以rmq的数组乘2就好了

&代码：

#include <cstdio>
#include <bitset>
#include <iostream>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)

const int max_v = 40000 + 7;
int rq[max_v*2][20], mm[max_v*2], n, m;

void initRMQ(int n, int b[]) {
   mm[0] = -1;
   for(int i = 1; i <= n; i++) {
      mm[i] = (i & (i - 1)) ? mm[i - 1] : mm[i - 1] + 1;
      rq[i][0] = i;
   }
   for(int j = 1; j <= mm[n]; j++)
      for(int i = 1; i + (1 << j) - 1 <= n; i++) {
         int x = rq[i][j - 1], y = rq[i + (1 << (j - 1))][j - 1];
         rq[i][j] = b[x] < b[y] ? x : y;
      }
}

int RMQ(int x, int y, int b[]) {
   if(x > y) swap(x, y);
   int k = mm[y - x + 1];
   int x2 = rq[x][k], y2 = rq[y - (1 << k) + 1][k];
   return b[x2] < b[y2] ? x2 : y2;
}

int dis[max_v];
struct Edge {
   int u, v, w, nx;
} G[2 * max_v];
int root;
int vs[max_v * 2];
int depth[max_v * 2];
int id[max_v];

int cnt, ft[max_v * 2], vis[max_v];

void Add(int u, int v, int w) {
   G[cnt].u = u; G[cnt].v = v; G[cnt].w = w;
   G[cnt].nx = ft[u];
   ft[u] = cnt++;
}

void dfs(int v, int p, int d, int &k) {
   id[v] = k;
   vs[k] = v;
   depth[k++] = d;
   // vis[v] = 1;
   for(int i = ft[v]; ~i; i = G[i].nx) {
      // printf("%d  %d--   %d
", G[i].u, G[i].v, p);
      if(G[i].v != p) {
         dis[G[i].v] = dis[G[i].u] + G[i].w;
         dfs(G[i].v, G[i].u, d + 1, k);
         vs[k] = v;
         depth[k++] = d;
      }
   }
}

void initLCA() {
   int k = 1;
   memset(vis, 0, sizeof(vis));
   dfs(root, -1, 0, k);
   initRMQ(k - 1, depth);
}

int LCA(int u, int v) {
   return vs[RMQ(min(id[u], id[v]), max(id[u], id[v]) + 1, depth)];
}

int main() {
   //("E:1.in", "r", stdin);
   int T; scanf("%d", &T);
   while(T--) {
      scanf("%d%d", &n, &m);
      cnt = 0;
      memset(ft, -1, sizeof(ft));
      fo(i, 1, n - 1) {
         int x, y, val;
         scanf("%d%d%d", &x, &y, &val);
         Add(x, y, val);
         Add(y, x, val);
      }
      root = 1;
      dis[root] = 0;
      initLCA();
      fo(i, 1, m) {
         int u, v;
         scanf("%d%d", &u, &v);
         printf("%d
", dis[u] + dis[v] - 2 * dis[LCA(u, v)]);
      }
   }
   return 0;
}

这个是用临街矩阵写的,RT了,但有注释

#include <cstdio>
#include <bitset>
#include <iostream>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)
#define pii pair<int,int>

//RMQ 相关
const int maxn = 40000 + 7, max_v = maxn;
int dp[maxn][20], mm[maxn], n, m;

void initRMQ(int n, int b[]) {
	mm[0] = -1;
	for(int i = 1; i <= n; i++) {
		mm[i] = (i & (i - 1)) ? mm[i - 1] : mm[i - 1] + 1;
		dp[i][0] = i;
	}
	for(int j = 1; j <= mm[n]; j++)
		for(int i = 1; i + (1 << j) - 1 <= n; i++) {
			int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1];
			dp[i][j] = b[x] < b[y] ? x : y;
			// dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
		}
}

int RMQ(int x, int y, int b[]) {
	if(x > y) swap(x, y);
	int k = mm[y - x + 1];
	int x2 = dp[x][k], y2 = dp[y - (1 << k) + 1][k];
	return b[x2] < b[y2] ? x2 : y2;
}


int dis[maxn];
//LCA 相关
vector<pii> G[max_v]; //图的邻接表
int root;
int vs[max_v * 2]; //dfs访问的顺序
int depth[max_v * 2]; //节点深度
int id[max_v]; //各顶点在vs中首次出现的下标

//v:现节点 p:父节点 d:深度 k:编号
void dfs(int v, int p, int d, int &k) {
	id[v] = k;
	vs[k] = v;
	depth[k++] = d;
	for(int i = 0; i < G[v].size(); i++) {
		if(G[v][i].first != p) {
			dis[G[v][i].first] = dis[v] + G[v][i].second;
			dfs(G[v][i].first, v, d + 1, k);
			vs[k] = v;
			depth[k++] = d;
		}
	}
}

void initLCA(int V) {
	int k = 1;
	dfs(root, -1, 0, k);
	initRMQ(k - 1, depth);
}

int LCA(int u, int v) {
	return vs[RMQ(min(id[u], id[v]), max(id[u], id[v]) + 1, depth)];
}


int main() {
	//("E:1.in", "r", stdin);
	int T; scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n, &m);
		fo(i, 1, n) G[i].clear();
		fo(i, 1, n - 1) {
			int x, y, val;
			scanf("%d%d%d", &x, &y, &val);
			G[x].push_back(make_pair(y, val));
			G[y].push_back(make_pair(x, val));
		}
		root = 1;
		dis[root] = 0;
		initLCA(n);
		fo(i, 1, m) {
			int u, v;
			scanf("%d%d", &u, &v);
			printf("%d
", dis[u] + dis[v] - 2 * dis[LCA(u, v)]);
		}
		// for(int i = 0; i < 8; i++) {
		// 	printf("%3d", i);
		// } printf("
");
		// for(int i = 0; i < 8; i++) {
		// 	printf("%3d", dis[i]);
		// } printf("
");
	}
	return 0;
}