正題

P1505

---

題目大意

給你一棵樹讓你進行以下操作

1.修改一條邊的權值

2.把一條鏈的權值取反

3.查詢一條鏈的權值和

4.查詢一條鏈的邊權最大值

5.查詢一條鏈的邊權最小值

---

解題思路

把邊拆成點，然后就是lct模板了

---

code

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define N 400010 using namespace std; int n,q,x,y,z,top,d[N]; char c[10]; struct Tree {#define ls son[x][0]#define rs son[x][1]int p[N],q[N],w[N],s[N],g[N],gg[N],mx[N],mn[N],fa[N],son[N][2];bool NR(int x){return fa[x]&&(son[fa[x]][0]==x||son[fa[x]][1]==x);}bool IRS(int x){return son[fa[x]][1]==x;}void push_up(int x){gg[x]=g[x]|gg[ls]|gg[rs];mn[x]=100000;mx[x]=-100000;s[x]=s[ls]+s[rs]+w[x];if(g[x])mn[x]=mx[x]=w[x];if(ls&&gg[ls])mn[x]=min(mn[x],mn[ls]),mx[x]=max(mx[x],mx[ls]);if(rs&&gg[rs])mn[x]=min(mn[x],mn[rs]),mx[x]=max(mx[x],mx[rs]);return;}void pushr(int x){p[x]^=1;swap(ls,rs);return;}void pushR(int x){w[x]=-w[x];s[x]=-s[x];swap(mn[x],mx[x]);mn[x]=-mn[x];mx[x]=-mx[x];q[x]^=1;return;}void push_down(int x){if(p[x]){if(ls)pushr(ls);if(rs)pushr(rs);p[x]=0;}if(q[x]){if(ls)pushR(ls);if(rs)pushR(rs);q[x]=0;}return;}void rotate(int x){int y=fa[x],z=fa[y],k=IRS(x),g=son[x][!k];if(NR(y))son[z][IRS(y)]=x;if(g)fa[g]=y;fa[x]=z;fa[y]=x;son[x][!k]=y;son[y][k]=g;push_up(y);return;}void Splay(int x){int y=x;d[++top]=y;while(NR(y))y=fa[y],d[++top]=y;while(top)push_down(d[top--]);while(NR(x)){if(NR(fa[x])){if(IRS(x)==IRS(fa[x]))rotate(fa[x]);else rotate(x);}rotate(x);}push_up(x);return;}void access(int x){for(int y=0;x;x=fa[y=x])Splay(x),rs=y,push_up(x);return;}void make_root(int x){access(x);Splay(x);pushr(x);return;}void link(int x,int y){make_root(x);fa[x]=y;return;}void Split(int x,int y){make_root(x);access(y);Splay(y);return;} }T; int main() {scanf("%d",&n);for(int i=1;i<n;++i){scanf("%d%d%d",&x,&y,&z);x++,y++;T.w[i+n]=z;T.g[i+n]=1;T.link(x,i+n);T.link(i+n,y);}scanf("%d",&q);while(q--){scanf("%s%d%d",c+1,&x,&y);if (c[1]=='C'){T.Splay(x+n);T.w[x+n]=y;T.push_up(x+n);}else if(c[1]=='N'){x++,y++;T.Split(x,y);T.pushR(y);}else if(c[1]=='S'){x++,y++;T.Split(x,y);printf("%d\n",T.s[y]);}else if(c[1]=='M'&&c[2]=='A'){x++,y++;T.Split(x,y);printf("%d\n",T.mx[y]);}else{x++,y++;T.Split(x,y);printf("%d\n",T.mn[y]);}}return 0; }

總結

以上是生活随笔為你收集整理的【LCT】旅游（P1505）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。