暑假集训 || LCA && RMQ

Posted 2020-12-31 pinkglightning

 

 

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;
}