LCA定义为对于一颗树 树上两个点的最近公共祖先
一.Tarjan求LCA（离线方法
https://blog.csdn.net/lw277232240/article/details/77017517
二.倍增法求LCA
void dfs(int u, int f) { for(int i = 1; i <= 18; i++) if(deep[u] >= (1<<i)) fa[u][i] = fa[fa[u][i-1]][i-1]; for(int i = head[u];i;i = nxt[i]) { int v = l[i].t; if(v != f) { deep[v] = deep[u] + 1; dist[v] = dist[u] + l[i].d; fa[v][0] = u; dfs(v, u); } } } int lca(int x, int y) { if(deep[x] < deep[y]) swap(x, y); int delta = deep[x] - deep[y]; for(int i = 0; i <= 18; i++) if((1<<i) & delta) x = fa[x][i]; for(int i = 18; i >= 0; i--) if(fa[x][i] != fa[y][i]) { x = fa[x][i]; y = fa[y][i]; } if(x == y) return x; else return fa[x][0]; } LL getdis(int x, int y) { int z = lca(x, y); return dist[x] + dist[y] - 2 * dist[z]; }
可以用来求一棵树上两点之间的最短距离
例题：
Gym 101808K 思路题
题意：给一个有n个点，n条边的图，n为1e5，查询任两点间的最短距离
思路：n个点n-1条边的话就是树，这个图就比树多了一条边，把这条边拿出来考虑
任意两点间的最短路有两种情况，一是经过这条边，二是不经过
建图的时候不加这一条边
x和y的距离 经过这条边的话x->uu + ww + vv->y或x->vv + ww + uu->y
不经过的话直接求即可
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; typedef long long LL; const int SZ = 200010; const int INF = 1e9+10; int head[SZ], nxt[SZ], tot = 0, deep[SZ], fa[SZ][20]; int fab[SZ]; LL dist[SZ]; struct node { int t, d; }l[SZ]; void build(int f, int t, int d) { l[++tot].t = t; l[tot].d = d; nxt[tot] = head[f]; head[f] = tot; } int n; void dfs(int u, int f) { for(int i = 1; i <= 18; i++) if(deep[u] >= (1<<i)) fa[u][i] = fa[fa[u][i-1]][i-1]; for(int i = head[u];i;i = nxt[i]) { int v = l[i].t; if(v != f) { deep[v] = deep[u] + 1; dist[v] = dist[u] + l[i].d; fa[v][0] = u; dfs(v, u); } } } int lca(int x, int y) { if(deep[x] < deep[y]) swap(x, y); int delta = deep[x] - deep[y]; for(int i = 0; i <= 18; i++) if((1<<i) & delta) x = fa[x][i]; for(int i = 18; i >= 0; i--) if(fa[x][i] != fa[y][i]) { x = fa[x][i]; y = fa[y][i]; } if(x == y) return x; else return fa[x][0]; } LL getdis(int x, int y) { int z = lca(x, y); return dist[x] + dist[y] - 2 * dist[z]; } void init() { memset(head, 0, sizeof(head)); tot = 0; memset(deep, 0, sizeof(deep)); memset(fa, 0, sizeof(fa)); for(int i = 1; i <= n; i++) fab[i] = i; } int find(int x) { return x == fab[x] ? x : fab[x] = find(fab[x]); } int main() { int T; scanf("%d", &T); while(T--) { int q; scanf("%d %d", &n, &q); init(); int uu, vv, ww; for(int i = 0; i < n; i++) { int u, v, w; scanf("%d %d %d", &u, &v, &w); int fu = find(u), fv = find(v); if(fu == fv) uu = u, vv = v, ww = w; else { fab[fu] = fv; build(u, v, w); build(v, u, w); } } dist[1] = 0; dfs(1, 0); //printf("%d %d ", uu, vv); //for(int i = 1; i <= n; i++) printf("%lld ", dist[i]); while(q--) { int x, y; scanf("%d %d", &x, &y); LL dis1 = getdis(x, y); LL dis2 = min(getdis(x, uu) + ww + getdis(y, vv), getdis(x, vv) + ww + getdis(y, uu)); printf("%lld ", min(dis1, dis2)); } } return 0; }
Gym 101810M
题意：一棵树，每条边来回都可以获得不同的值，每条边只能去一次回一次，任意查询从s到t最多能获得多少值
思路：发现能获得的值是整棵树上的权值 - 从t走最短路到s获得的值
用dist[0][u]记录从根走到u总的值
用dist[1][u]记录从u走到根总的值
画个图推个式子就ok了
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; typedef long long LL; const int SZ = 400010; const int INF = 1e9+10; int head[SZ], nxt[SZ], tot = 0, deep[SZ], fa[SZ][20]; int dist[2][SZ]; struct node { int t, c1, c2; }l[SZ]; void build(int f, int t, int c1, int c2) { l[++tot].t = t; l[tot].c1 = c1; l[tot].c2 = c2; nxt[tot] = head[f]; head[f] = tot; } int n; void dfs(int u, int f) { for(int i = 1; i <= 18; i++) if(deep[u] >= (1<<i)) fa[u][i] = fa[fa[u][i-1]][i-1]; for(int i = head[u];i;i = nxt[i]) { int v = l[i].t; if(v != f) { deep[v] = deep[u] + 1; dist[0][v] = dist[0][u] + l[i].c1; dist[1][v] = dist[1][u] + l[i].c2; fa[v][0] = u; dfs(v, u); } } } int lca(int x, int y) { if(deep[x] < deep[y]) swap(x, y); int delta = deep[x] - deep[y]; for(int i = 0; i <= 18; i++) if((1<<i) & delta) x = fa[x][i]; for(int i = 18; i >= 0; i--) if(fa[x][i] != fa[y][i]) { x = fa[x][i]; y = fa[y][i]; } if(x == y) return x; else return fa[x][0]; } void init() { memset(head, 0, sizeof(head)); tot = 0; memset(deep, 0, sizeof(deep)); memset(fa, 0, sizeof(fa)); } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d", &n); init(); int sum = 0; for(int i = 0; i < n-1; i++) { int u, v, w1, w2; scanf("%d %d %d %d", &u, &v, &w1, &w2); build(u, v, w1, w2); build(v, u, w2, w1); sum = sum + w1 + w2; } dist[0][1] = 0, dist[1][1] = 0; dfs(1, 0); int q; scanf("%d", &q); while(q--) { int x, y; scanf("%d %d", &x, &y); int z = lca(x, y); int ans = dist[1][y] - dist[1][z] + dist[0][x] - dist[0][z]; printf("%d ", sum - ans); } } return 0; }
RMQ：区间最值查询问题
用f[i][j]表示 从a[i] 开始 往后2^j个数里面的 最大/最小 值或GCD
void st(int n) { for(int i = 1; i <= n; i++) f[i][0] = a[i]; for(int j = 1; (1 << j) <= n; j++) for(int i = 1; i + (1 << j) - 1 <= n; i++) f[i][j] = max(f[i][j-1], f[i + (1<<(j-1))][j-1]); } int RMQ(int l, int r) { int k = 0; while((1<<(k + 1) <= r - l + 1)) k++; return max(f[l][k], f[r - (1<<k) + 1][k]); }
最小值把max改成min， GCD把max改成__gcd
例题：
POJ 2019 二维RMQ
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; const int SZ = 550; const int INF = 1e9+10; int a[SZ][SZ], mmax[SZ][SZ][15], mmin[SZ][SZ][15]; void st(int n) { for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) mmax[i][j][0] = mmin[i][j][0] = a[i][j]; for(int j = 1; (1 << j) <= n; j++) for(int i = 1; i + (1 << j) - 1 <= n; i++) for(int k = 1; k <= n; k++) { mmax[k][i][j] = max(mmax[k][i][j-1], mmax[k][i + (1<<(j-1))][j-1]); mmin[k][i][j] = min(mmin[k][i][j-1], mmin[k][i + (1<<(j-1))][j-1]); } } int RMQ(int x, int l, int r, int b) { int k = 0; while((1<<(k + 1) <= r - l + 1)) k++; int ans_max = -INF, ans_min = INF; for(int i = x; i < x + b; i++) { ans_max = max(ans_max, max(mmax[i][l][k], mmax[i][r- (1<<k) + 1][k])); ans_min = min(ans_min, min(mmin[i][l][k], mmin[i][r- (1<<k) + 1][k])); } return (ans_max - ans_min); } int main() { int n, b, k; scanf("%d %d %d", &n, &b, &k); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) scanf("%d", &a[i][j]); st(n); for(int i = 0; i < k; i++) { int x, y; scanf("%d %d", &x, &y); int l = y, r = y + b - 1; printf("%d ", RMQ(x, l, r, b)); } return 0; }
HDU 5726
题意：给一段序列，任意查询一个区间内所有数的GCD，以及有多少个区间的GCD数和它相同
思路：RMQ+二分。。考场上有思路了但是没敢敲QAQ
发现序列越长，GCD是不增的，于是对于每个数可以通过二分判断它往后多少个数，这一段里面拥有相同的GCD
用map记录即可
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <map> using namespace std; typedef long long LL; const int SZ = 100010; const int INF = 1e9+10; int a[SZ]; int f[SZ][22]; map<int, LL> mp; void st(int n) { for(int i = 1; i <= n; i++) f[i][0] = a[i]; for(int j = 1; (1 << j) <= n; j++) for(int i = 1; i + (1 << j) - 1 <= n; i++) { f[i][j] = __gcd(f[i][j-1], f[i + (1<<(j-1))][j-1]); } } int RMQ(int l, int r) { int k = 0; while((1<<(k + 1) <= r - l + 1)) k++; return __gcd(f[l][k], f[r - (1<<k) + 1][k]); } int main() { int T, tt = 0; scanf("%d", &T); while(T--) { int n; scanf("%d", &n); mp.clear(); for(int i = 1; i <= n; i++) for(int j = 1; j <= 18; j++) f[i][j] = 1; for(int i = 1; i <= n; i++) scanf("%d", &a[i]); st(n); for(int i = 1; i <= n; i++) { int k = i; while(k <= n) { int l = k, r = n; while(l <= r) { int mid = (l + r + 1) >> 1; if(RMQ(i, mid) < RMQ(i, k)) r = mid - 1; else l = mid + 1; } mp[RMQ(i, k)] += LL(l - k); k = l; } } int q; scanf("%d", &q); printf("Case #%d: ", ++tt); while(q--) { int x, y; scanf("%d %d", &x, &y); int ans = RMQ(x, y); printf("%d %lld ", ans, mp[ans]); } } return 0; }