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【BZOJ2631】tree (LCT)

时间:2016-01-18 19:08:16      阅读:227      评论:0      收藏:0      [点我收藏+]

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链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2631

终于学了LCT惹qwq (看的板是黄学长的orzzz

T了很久发现窝开了个ch[MaxN][0] ,这都能编译过?

 

技术分享
  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <iostream>
  4 #include <algorithm>
  5 #define mod 51061
  6 #define MaxN 100010
  7 using namespace std;
  8 int n, q;
  9 int ch[MaxN][2]; 
 10 unsigned int sum[MaxN], sz[MaxN], ad[MaxN], mu[MaxN], Q[MaxN], p[MaxN], flip[MaxN], val[MaxN];
 11 
 12 void updata(int x, int m, int a){
 13     val[x] = (val[x]*m + a) % mod;
 14     sum[x] = (sum[x]*m + a*sz[x]) % mod; 
 15     ad[x] = (ad[x] * m + a) % mod;
 16     mu[x] = (mu[x] * m) % mod;
 17 }
 18     
 19 void push_up(int x){
 20     int l = ch[x][0], r = ch[x][1];
 21     sz[x] = (sz[l]+sz[r]+1) % mod;
 22     sum[x] = (sum[l]+sum[r]+val[x]) % mod;
 23 }
 24 
 25 void push_down(int x){
 26     int l = ch[x][0], r = ch[x][1];
 27         if (flip[x]) {
 28         flip[x] = 0; flip[l] ^= 1; flip[r] ^= 1;
 29         swap(ch[x][0], ch[x][1]);
 30     }
 31     int m = mu[x], a = ad[x];
 32     if (m != 1 || a != 0) {
 33         updata(l, m, a);
 34         updata(r, m, a);
 35         mu[x] = 1, ad[x] = 0;
 36     }
 37 }
 38     
 39 bool isroot(int x){
 40     return ch[p[x]][0] != x && ch[p[x]][1] != x;
 41 }
 42     
 43 void rotate(int x){
 44     int y = p[x], z = p[y];
 45     if (!isroot(y)) ch[z][ch[z][1] == y] = x;
 46     int l = ch[y][1] == x, r = l ^ 1;
 47     p[y] = x;
 48     p[x] = z;
 49     p[ch[x][r]] = y;
 50     
 51     ch[y][l] = ch[x][r];
 52     ch[x][r] = y;
 53     push_up(y);
 54     push_up(x);
 55 }
 56 
 57 void splay(int x){
 58     int top = 1;
 59     Q[top] = x;
 60     for (int t = x; !isroot(t); t = p[t]) Q[++top] = p[t];
 61     while (top) push_down(Q[top--]);
 62     while (!isroot(x)){
 63         int y = p[x], z = p[y];
 64         if (!isroot(y)){
 65             if ((ch[y][1] == x) ^ (ch[z][1] == y)) rotate(x);
 66             else rotate(y);
 67         }
 68         rotate(x);
 69     }
 70 }
 71 
 72 void Access(int x){
 73     for (int t = 0; x; t = x, x = p[x]){
 74         splay(x); ch[x][1] = t; push_up(x);            
 75     }
 76 }
 77     
 78 void Make_root(int x){
 79     Access(x); splay(x); flip[x] ^= 1;
 80 }
 81 
 82 void Link(int u, int v){
 83     Make_root(u); p[u] = v;
 84 }    
 85 
 86 void Cut(int u, int v){
 87     Make_root(u); Access(v); splay(v);
 88     p[u] = ch[v][0] = 0;
 89 }
 90 
 91 void Split(int u, int v){
 92     Make_root(v); Access(u); splay(u);
 93 }
 94 
 95 void Read_Data(){
 96     scanf("%d%d", &n, &q);
 97     for (int i = 1; i <= n; i++) 
 98         sum[i] = sz[i] = mu[i] = val[i] = 1;
 99     for (int i = 1, u, v; i < n; i++){
100         scanf("%d%d", &u, &v);
101         Link(u, v);
102     }
103 }
104 
105 void Solve(){
106     char s[5];
107     int u, v, c, u2, v2;
108     for (int i = 1; i <= q; i++){
109         scanf("%s%d%d", s, &u, &v);
110         if (s[0] == +) {
111             scanf("%d", &c);
112             Split(u, v); updata(u, 1, c);
113         } 
114         if (s[0] == -) {
115             scanf("%d%d", &u2, &v2);
116             Cut(u, v); Link(u2, v2);
117         }
118         if (s[0] == *) {
119             scanf("%d", &c);
120             Split(u, v); updata(u, c, 0);    
121         }
122         if (s[0] == /) {
123             Split(u, v); 
124             printf("%d\n", sum[u]);
125         }
126     }
127 }
128 
129 int main(){
130     freopen("bzoj2631.in", "r", stdin);
131     freopen("bzoj2631.out", "w", stdout);
132     Read_Data();
133     Solve();
134     return 0;    
135 }
_(:з」∠)_

 

【BZOJ2631】tree (LCT)

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原文地址:http://www.cnblogs.com/Lukaluka/p/5139984.html

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