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LA - 5031 - Graph and Queries

2013-09-10 20:16 232 查看
题意:一个N个点(编号从1开始),M条边的无向图(编号从1开始),有3种操作:

D X:把编号为X的边删了;

Q X K:查询编号为X的结点所在连通分量第K大的元素;

C X V:将编号为X的结点的权值修改为V。

问所有查询的结果的平均值(1 <= N <= 20000, 0 <= M <= 60000, -10^6 <= 点权 <= 10^6, 1 <= Q操作次数 <= 2 * 10^5, C操作次数 <= 2 * 10^5)。

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3032

——>>LJ《训练指南》Treap树的例题,题目中关于Q操作的话:among all vertexes currently connected with vertex X,总觉得指的是与X直接相连的结点,可实现上却应理解为X所在连通分量的所有结点。

#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

const int maxn = 20000 + 10;
const int maxm = 60000 + 10;
const int maxc = 500000 + 10;

struct Command{
char type;
int x, p;
Command(char type = '\0', int x = 0, int p = 0):type(type), x(x), p(p){}
};

struct Node{
Node *ch[2];
int r;
int v;
int s;
Node(int v):v(v){
ch[0] = ch[1] = NULL;
r = rand();
s = 1;
}

bool operator < (const Node& e) const{
return r < e.r;
}

int cmp(int x) const{
if(x == v) return -1;
return x < v ? 0 : 1;
}

void maintain(){
s = 1;
if(ch[0] != NULL) s += ch[0]->s;
if(ch[1] != NULL) s += ch[1]->s;
}
};

int N, M, weight[maxn], from[maxm], to[maxm], fa[maxn], kase, c, query_cnt;
long long query_tot;
bool removed[maxm];
Command commands[maxc];
Node *root[maxn];

int Find(int x){
return x == fa[x] ? x : Find(fa[x]);
}

void removetree(Node* &x){
if(x->ch[0] != NULL) removetree(x->ch[0]);
if(x->ch[1] != NULL) removetree(x->ch[1]);
delete x;
x = NULL;
}

void rotate(Node* &o, int d){
Node* k = o->ch[d^1];
o->ch[d^1] = k->ch[d];
k->ch[d] = o;
o->maintain();
k->maintain();
o = k;
}

void insert(Node* &o, int x){
if(o == NULL) o = new Node(x);
else{
int d = x < o->v ? 0 : 1;
insert(o->ch[d], x);
if(o->ch[d] > o) rotate(o, d^1);
}
o->maintain();
}

void remove(Node* &o, int x){
int d = o->cmp(x);
if(d == -1){
Node* u = o;
if(o->ch[0] != NULL && o->ch[1] != NULL){
int d2 = o->ch[0] > o->ch[1] ? 1 : 0;
rotate(o, d2);
remove(o->ch[d2], x);
}
else{
if(o->ch[0] == NULL) o = o->ch[1];
else o = o->ch[0];
delete u;
}
}
else remove(o->ch[d], x);
if(o != NULL) o->maintain();
}

void mergeto(Node* &src, Node* &dest){
if(src->ch[0] != NULL) mergeto(src->ch[0], dest);
if(src->ch[1] != NULL) mergeto(src->ch[1], dest);
insert(dest, src->v);
delete src;
src = NULL;
}

void addEdge(int x){
int u = Find(from[x]);
int v = Find(to[x]);
if(u != v){
if(root[u]->s < root[v]->s){
fa[u] = v;
mergeto(root[u], root[v]);
}
else{
fa[v] = u;
mergeto(root[v], root[u]);
}
}
}

int kth(Node* o, int k){
if(o == NULL || k <= 0 || k > o->s) return 0;
int s = (o->ch[1] == NULL ? 0 : o->ch[1]->s);
if(k == s+1) return o->v;
else if(k < s+1) return kth(o->ch[1], k);
else return kth(o->ch[0], k-s-1);
}

void query(int x, int k){
query_cnt++;
query_tot += kth(root[Find(x)], k);
}

void change_weight(int x, int v){
int u = Find(x);
remove(root[u], weight[x]);
insert(root[u], v);
weight[x] = v;
}

void read(){
for(int i = 1; i <= N; i++) scanf("%d", &weight[i]);
for(int i = 1; i <= M; i++) scanf("%d%d", &from[i], &to[i]);
c = 0;
memset(removed, 0, sizeof(removed));
while(1){
char type;
int X, K, V;
scanf(" %c", &type);
if(type == 'E') break;
scanf("%d", &X);
if(type == 'D') removed[X] = 1;
else if(type == 'Q') scanf("%d", &K);
else{
scanf("%d", &V);
K = weight[X];
weight[X] = V;
}
commands[c++] = Command(type, X, K);
}
}

void build(){
for(int i = 1; i <= N; i++){
fa[i] = i;
if(root[i] != NULL) removetree(root[i]);
root[i] = new Node(weight[i]);
}
for(int i = 1; i <= M; i++) if(!removed[i]) addEdge(i);
}

void solve(){
query_tot = query_cnt = 0;
for(int i = c-1; i >= 0; i--){
if(commands[i].type == 'D') addEdge(commands[i].x);
else if(commands[i].type == 'Q') query(commands[i].x, commands[i].p);
else change_weight(commands[i].x, commands[i].p);
}
printf("Case %d: %.6lf\n", kase++, (double)query_tot / query_cnt);
}

int main()
{
kase = 1;
while(scanf("%d%d", &N, &M) == 2){
if(!N && !M) return 0;
read();
build();
solve();
}
return 0;
}
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