今天闲的搜了一下.cpp...
神奇的Windows告诉我有499个文件;
吓得我赶紧写了一道水题...就500啦
虽然不多但是开心(^_^)
(附上树链表示庆祝)
树链剖分求LCA
复杂度忘了但是是本菜鸡已知的唯一能过luoguP3379的 = =
首先普及一下链(重链)
重链就是每次都从下两个儿子中选择较大的一个接上...
往下子孙的个数多的叫大儿子(回溯...)
根是第一个起点，找完一条链就从剩下的中选择深度dep最小的继续
然鹅有什么用??
求解过程:
①输入x,y
②求出链顶(起点)
③判断是否相同
如果相同:
LCA就是深度较小的那个(不解释),break
如果不同:
选择深度较深的往上走一步(省时啊!!)
⑤继续判断
⑨结束
②③覆盖x,y值
代码贴上(read()快读可以用标输)
18.09.21 更新
很久以前的blog……感慨颇多啊
前两天在网上到处搜最后还是看了自己的博客重新学会的……
粘一下代码lo
luoguP3384
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #define rep(i,a,n) for(int i = a;i <= n;i++) 7 #define per(i,n,a) for(int i = n;i >= a;i--) 8 #define ls t<<1 9 #define rs t<<1|1 10 using namespace std; 11 typedef long long ll; 12 int read() { 13 int as = 0,fu = 1; 14 char c = getchar(); 15 while(c < '0' || c > '9') { 16 if(c == '-') fu = -1; 17 c = getchar(); 18 } 19 while(c >= '0' && c <= '9') { 20 as = as * 10 + c - '0'; 21 c = getchar(); 22 } 23 return as * fu; 24 } 25 26 const int N = 200005; 27 //head 28 29 int n,T,s,mod; 30 int cnt,cur,head[N],nxt[N],mo[N]; 31 int w[N],a[N],p[N<<2],add[N<<2]; 32 int id[N],pa[N],mson[N],top[N],sz[N],dep[N]; 33 34 void addline(int x,int y) { 35 mo[++cnt] = y; 36 nxt[cnt] = head[x]; 37 head[x] = cnt; 38 mo[++cnt] = x; 39 nxt[cnt] = head[y]; 40 head[y] = cnt; 41 return; 42 } 43 44 void dfs1(int x,int fa,int d) { 45 dep[x] = d; 46 pa[x] = fa; 47 sz[x] = 1; 48 int maxson = -1; 49 for(int i = head[x]; i; i = nxt[i]) { 50 int sn = mo[i]; 51 if(sn == fa) continue; 52 dfs1(sn,x,d+1); 53 sz[x] += sz[sn]; 54 if(sz[sn] > maxson) { 55 mson[x] = sn; 56 maxson = sz[sn]; 57 } 58 } 59 } 60 void dfs2(int x,int topx) { 61 id[x] = ++cur; 62 a[cur] = w[x]; 63 top[x] = topx; 64 if(mson[x] == 0) return; 65 dfs2(mson[x],topx); 66 for(int i = head[x]; i; i = nxt[i]) { 67 int sn = mo[i]; 68 if(sn == mson[x] || sn == pa[x]) continue; 69 dfs2(sn,sn); 70 } 71 return; 72 } 73 74 inline void pup(int t) { 75 p[t] = (p[ls] + p[rs]) % mod; 76 } 77 inline void pdown(int l,int r,int t) { 78 if(add[t] == 0) return; 79 int m = l+r >> 1; 80 int szl = m+1-l; 81 int szr = r-m; 82 p[ls] = (p[ls] + add[t] * szl) % mod; 83 p[rs] = (p[rs] + add[t] * szr) % mod; 84 add[ls] = (add[ls] + add[t]) % mod; 85 add[rs] = (add[rs] + add[t]) % mod; 86 add[t] = 0; 87 return; 88 } 89 void build(int l,int r,int t) { 90 if(l == r) { 91 p[t] = a[l]; 92 return; 93 } 94 int m = l+r >>1; 95 build(l,m,ls); 96 build(m+1,r,rs); 97 pup(t); 98 return; 99 } 100 void Segupdate(int L,int R,int c,int l,int r,int t) { 101 if(L <= l && r <= R) { 102 p[t] = (p[t] + c * (r - l + 1)) % mod; 103 add[t] = (add[t] + c) % mod; 104 return; 105 } 106 pdown(l,r,t); 107 int m = l+r >> 1; 108 if(L <= m) Segupdate(L,R,c,l,m,ls); 109 if(R > m) Segupdate(L,R,c,m+1,r,rs); 110 pup(t); 111 return; 112 } 113 int Segquery(int L,int R,int l,int r,int t) { 114 if(L <= l && r <= R) return p[t]; 115 pdown(l,r,t); 116 int m = l+r >> 1; 117 int ans = 0; 118 if(L <= m) ans = (ans + Segquery(L,R,l,m,ls)) % mod; 119 if(R > m) ans = (ans + Segquery(L,R,m+1,r,rs)) % mod; 120 return ans; 121 } 122 123 void Lineupdate(int x,int y,int z) { 124 while(top[x] != top[y]) { 125 if(dep[top[x]] < dep[top[y]]) swap(x,y); 126 Segupdate(id[top[x]],id[x],z,1,n,1); 127 x = pa[top[x]]; 128 } 129 if(dep[x] < dep[y]) swap(x,y); 130 Segupdate(id[y],id[x],z,1,n,1); 131 return; 132 } 133 int Linequery(int x,int y) { 134 int ans = 0; 135 while(top[x] != top[y]) { 136 if(dep[top[x]] < dep[top[y]]) swap(x,y); 137 ans = (ans + Segquery(id[top[x]],id[x],1,n,1)) % mod; 138 x = pa[top[x]]; 139 } 140 if(dep[x] < dep[y]) swap(x,y); 141 ans = (ans + Segquery(id[y],id[x],1,n,1)) % mod; 142 return ans; 143 } 144 145 void Sonupdate(int x,int z) { 146 Segupdate(id[x],id[x] + sz[x] - 1,z,1,n,1); 147 return; 148 } 149 int Sonquery(int x) { 150 return Segquery(id[x],id[x] + sz[x] - 1,1,n,1); 151 } 152 int main() { 153 n = read(),T = read(),s = read(),mod = read(); 154 rep(i,1,n) w[i] = read() % mod; 155 rep(i,2,n) addline(read(),read()); 156 dfs1(s,0,0); 157 dfs2(s,s); 158 build(1,n,1); 159 while(T--) { 160 int op = read(); 161 if(op == 1) { 162 int x = read(); 163 int y = read(); 164 int z = read() % mod; 165 Lineupdate(x,y,z); 166 } 167 if(op == 2) { 168 int x = read(); 169 int y = read(); 170 printf("%d ",Linequery(x,y)); 171 } 172 if(op == 3) { 173 int x = read(); 174 int z = read() % mod; 175 Sonupdate(x,z); 176 177 } 178 if(op == 4) { 179 int x = read(); 180 printf("%d ",Sonquery(x)); 181 } 182 } 183 return 0; 184 }
有点小小的忧伤啊……