Vijos上的三道LCA: 1427, 1460, 1710

Vijos上一共有三道标记为LCA的题目：P1427机密信息，P1460拉力赛，P1710Mrw的工资计划。

首先是P1427机密信息。考虑到只需要求一次LCA，数据范围也不大，直接暴力解决，只是分类讨论有点麻烦。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define rep(i,l,r) for(int i=l; i<=r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(int i=last[x]; i; i=edge[i].pre)
typedef long long ll;
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 10010;
struct Edge{
    int pre,to,cost;
}edge[maxn];
int n,s,t,x,y,tot=0,root,totaltime=0,d[maxn],last[maxn],depth[maxn],father[maxn],countt[maxn],timee[maxn];
string str;
inline int read(){
    int ans = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)){
        if (c == '-') f = -1;
        c = getchar();
    }
    while (isdigit(c)){
        ans = ans * 10 + c - '0';
        c = getchar();
    }
    return ans * f;
}
inline void addedge(int x,int y,int z){
    edge[++tot].pre = last[x];
    edge[tot].to = y;
    edge[tot].cost = z;
    last[x] = tot;
}
void dfs(int x){
    travel(x){
        depth[edge[i].to] = depth[x] + 1;
        timee[edge[i].to] = timee[x] + countt[edge[i].to];
        d[edge[i].to] = d[x] + edge[i].cost;
        dfs(edge[i].to);
    }
}
int lca(int x,int y){
    if (depth[x] < depth[y]) swap(x,y);
    while (depth[x] != depth[y]) x = father[x];
    while (x != y){
        x = father[x]; y = father[y];
    }
    return x;
}
void solve(){
    int l = lca(s,t);
    int res = d[s] + d[t] - d[l] - d[father[l]];
    printf("%d
",res);
    if (l != s && l != t) totaltime = timee[father[s]] + timee[father[t]] - timee[l] - timee[father[l]];
    else if (l == s) totaltime = timee[father[t]] - timee[s];
    else if (l == t) totaltime = timee[father[s]] - timee[father[t]];
    printf("%d
",totaltime);
}
int main(){
    n = read(); s = read(); t = read(); clr(countt,0); clr(last,0);
    rep(i,1,n){
        x = read(); y = read(); cin >> str;
        addedge(y,x,str.length());
        father[x] = y; countt[y]++;
    }
    d[0] = 0; dfs(0);
    solve();
    return 0;
}

P1460拉力赛，非常裸的一道倍增求LCA。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define rep(i,l,r) for (int i=l; i<=r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for (int i=last[x]; i; i=edge[i].pre)
typedef long long ll;
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 10010;
struct Edge{
    ll pre,toward,cost;
}edge[maxn<<1];
ll n,m,x,y,z,now,tot=0,total,totald,father[maxn],last[maxn],fa[maxn][16],depth[maxn],d[maxn];
bool vis[maxn];
inline ll read(){
    int ans = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)){
        if (c== '-') f = -1;
        c = getchar();
    }
    while (isdigit(c)){
        ans = ans * 10 + c - '0';
        c = getchar();
    }
    return ans * f;
}
inline void addedge(ll x,ll y,ll z){
    edge[++tot].pre = last[x];
    edge[tot].toward = y;
    edge[tot].cost = z;
    last[x] = tot;
}
void bfs(){
    queue <ll> q; clr(vis,0);
    q.push(1); depth[1] = 0; d[1] = 0;
    while (!q.empty()){
        now = q.front();
        q.pop();
        if (vis[now]) continue;
        vis[now] = 1;
        rep(i,1,16){
            if (depth[now] < (1 << i)) break;
            fa[now][i] = fa[fa[now][i-1]][i-1];
        }
        travel(now){
            if (!vis[edge[i].toward]){
                depth[edge[i].toward] = depth[now] + 1;
                d[edge[i].toward] = d[now] + edge[i].cost;
                fa[edge[i].toward][0] = now;
                q.push(edge[i].toward);
            }
        }
    }
}
ll lca(ll x,ll y){
    if (depth[x] < depth[y]) swap(x,y);
    int t = depth[x] - depth[y];
    rep(i,0,16) if (t & (1 << i)) x = fa[x][i];
    for (int i=16; i>=0; i--){
        if (fa[x][i] != fa[y][i]){
            x = fa[x][i];
            y = fa[y][i];
        }
    }
    if (x == y) return x;
    return fa[x][0];
}
int main(){
    n = read(); m = read();
    rep(i,1,n-1){
        x = read(); y = read(); z = read();
        addedge(x,y,z); addedge(y,x,z);
    }
    bfs();
    rep(i,1,m){
        x = read(); y = read();
        if (lca(x,y) == x) total++, totald += d[y] - d[x];
    }
    printf("%I64d
%I64d
",total,totald);
    return 0;
}

P1710Mrw的工资计划，倍增LCA加树状数组。

题目询问奇数层与偶数层的工资之差的绝对值，将奇数层的工资存为正数，将偶数层存为负数，然后用树状数组记录每一层的增加值，计算时加上即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define rep(i,l,r) for(int i=l; i<=r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(int i=last[x]; i; i=edge[i].pre)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 20010;
struct Edge{
    int pre,to;
}edge[maxn];
int n,lv,tot=0,t,q,x,y,boss,maxdep=-1,last[maxn],tree[maxn],sum[maxn],pay[maxn],fa[maxn][14],depth[maxn];
char ch;
bool vis[maxn];
inline int read(){
    int ans = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)){
        if (c == '-') f = -1;
        c = getchar();
    }
    while (isdigit(c)){
        ans = ans * 10 + c - '0';
        c = getchar();
    }
    return ans * f;
}
inline int lowbit(int x){
    return x & (-x);
}
inline void addedge(int x,int y){
    edge[++tot].pre = last[x];
    edge[tot].to = y;
    last[x] = tot;
}
void add(int x,int t){
    for(int i=x; i<=maxdep; i+=lowbit(i))
    tree[i] += t;
}
int query(int x){
    int ret = 0;
    for(int i=x; i; i-=lowbit(i))
    ret += tree[i];
    return ret;
}
void dfs(int x){
    vis[x] = 1;
    rep(i,1,13){
        if (depth[x] < (1 << i)) break;
        fa[x][i] = fa[fa[x][i-1]][i-1];
    }
    travel(x){
        if (vis[edge[i].to]) continue;
        vis[edge[i].to] = 1;
        depth[edge[i].to] = depth[x] + 1;
        if (!(depth[edge[i].to] & 1)) pay[edge[i].to] = -pay[edge[i].to];
        sum[edge[i].to] = sum[x] + pay[edge[i].to];
        maxdep = max(maxdep,depth[edge[i].to]);
        fa[edge[i].to][0] = x;
        dfs(edge[i].to);
    }
}
int lca(int x,int y){
    if (depth[x] < depth[y]) swap(x,y);
    int t = depth[x] - depth[y];
    rep(i,0,13) if (t & (1 << i)) x = fa[x][i];
    for(int i=13; i>=0; i--) if (fa[x][i] != fa[y][i]){
        x = fa[x][i]; y = fa[y][i];
    }
    if (x == y) return x; return fa[x][0];
}
int main(){
    n = read(); q = read(); clr(last,0);
    rep(i,1,n) pay[i] = read();
    rep(i,1,n){
        x = read();
        if (x == -1) boss = i;
        else addedge(x,i);
    }
    clr(vis,0); depth[boss] = 1; dfs(boss); clr(tree,0);
    rep(i,1,q){
        scanf("%c",&ch);
        if (ch == 'M'){
            lv = read(); t = read();
            if (lv & 1) add(lv,t); else add(lv,(-t));
        }
        else{
            x = read(); y = read();
            int l = lca(x,y);
            printf("%d
",abs(sum[x] + sum[y] - (sum[l] << 1) + pay[l] + query(depth[x]) + query(depth[y])
            - query(depth[l]) - query(depth[l] - 1)));
        }
    }
    return 0;
}