zoukankan      html  css  js  c++  java
  • Bzoj2959: 长跑

    题面

    传送门

    Sol

    (LCT+)并查集维护双联通分量,每次加边(x,y),如果已经相连就把这条路径缩成一个点,赋上权值
    (LCT)调用(fa)时一定要在并查集中(find)一下
    细节很多,常数很大

    # include <bits/stdc++.h>
    # define IL inline
    # define RG register
    # define ls ch[0][x]
    # define rs ch[1][x]
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(2e5 + 10);
    
    IL ll Read(){
        RG char c = getchar(); RG ll x = 0, z = 1;
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    int n, Q, ch[2][_], fa[_], sum[_], rev[_], S[_], val[_], scc[_], bel[_], w[_];
    
    IL int Find(RG int x){  return x == scc[x] ? x : scc[x] = Find(scc[x]);  }
    
    IL int _Find(RG int x){  return x == bel[x] ? x : bel[x] = _Find(bel[x]);  }
    
    IL bool Son(RG int x){  return ch[1][Find(fa[x])] == x;  }
    
    IL bool Isroot(RG int x){  RG int pa = Find(fa[x]); return ch[0][pa] != x && ch[1][pa] != x;  }
    
    IL void Reverse(RG int x){  if(!x) return; swap(ls, rs); rev[x] ^= 1;  }
    
    IL void Update(RG int x){  sum[x] = sum[ls] + sum[rs] + val[x];  }
    
    IL void Pushdown(RG int x){  if(!rev[x]) return; Reverse(ls); Reverse(rs); rev[x] = 0;  }
    
    IL void Rotate(RG int x){
        RG int y = Find(fa[x]), z = Find(fa[y]), c = Son(x);
        if(!Isroot(y)) ch[Son(y)][z] = x; fa[x] = z;
        ch[c][y] = ch[!c][x]; fa[ch[c][y]] = y;
        ch[!c][x] = y; fa[y] = x; Update(y);
    }
    
    IL void Splay(RG int x){
        S[S[0] = 1] = x;
        for(RG int y = x; !Isroot(y); y = Find(fa[y])) S[++S[0]] = Find(fa[y]);
        while(S[0]) Pushdown(S[S[0]--]);
        for(RG int y = Find(fa[x]); !Isroot(x); Rotate(x), y = Find(fa[x]))
            if(!Isroot(y)) Son(x) ^ Son(y) ? Rotate(x) : Rotate(y);
        Update(x);
    }
    
    IL void Access(RG int x){  for(RG int y = 0; x; y = x, x = Find(fa[x])) Splay(x), ch[1][x] = y, Update(x);  }
    
    IL void Makeroot(RG int x){  Access(x); Splay(x); Reverse(x);  }
    
    IL void Link(RG int x, RG int y){  Makeroot(x); fa[x] = y;  }
    
    IL void Split(RG int x, RG int y){  Makeroot(x); Access(y); Splay(y);  }
    
    IL void Dfs(RG int x, RG int y){  if(!x) return; scc[Find(x)] = y; Dfs(ls, y); Dfs(rs, y);  }
    
    int main(RG int argc, RG char* argv[]){
        n = Read(); Q = Read();
    	for(RG int i = 1; i <= n; ++i) bel[i] = scc[i] = i, w[i] = val[i] = Read();
        while(Q--){
    		RG int p = Read(), a = Read(), b = Read(), fx;
    		if(p == 1){
    			a = Find(a); b = Find(b);
    			if(_Find(a) != _Find(b)) Link(a, b), bel[_Find(a)] = _Find(b);
    			else{
    				Split(a, b); val[b] = sum[b];
    				Dfs(b, b);
    				ch[0][b] = ch[1][b] = 0;
    			}
    		}
    		else if(p == 2) fx = Find(a), Makeroot(fx), val[fx] += b - w[a], w[a] = b, Update(fx);
    		else{
    			a = Find(a); b = Find(b);
    			if(_Find(a) != _Find(b)){  puts("-1"); continue;  }
    			Split(a, b); printf("%d
    ", sum[b]);
    		}
        }
        return 0;
    }
    
    
  • 相关阅读:
    django 设置局域网内访问项目
    104. Maximum Depth of Binary Tree
    multi dimension array
    Longest Plindrome Subarray
    BlockQueue
    H2O
    HoppingIterator
    midStack
    Reverse Words in a String II
    public boolean canPlaceFlowers(List<Boolean> flowerbed, int numberToPlace)
  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8331356.html
Copyright © 2011-2022 走看看