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UVaLive 5031 Graph and Queries (Treap)
题意:初始时给出一个图,每个点有一个权值,三种操作:(1)删除某个边;(2)修改每个点的权值;(3)询问与节点x在一个连通分量中所有点的第K大的权值。
析:首先是要先离线,然后再倒着做,第一个操作就成了加边操作,很容易实现,第二操作,就是分成两个操作,先把x结点删掉,然后再插入一个新结点,
最后一个是就是求某个连通分量的第 k 大,直接用treap直接查找就好,注意问是第 k 大,不是第 k 小。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 5e5 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int w[maxn], a[maxn], b[maxn]; bool removed[maxn]; LL cnt, tot; struct Command{ char type; int x, p; }; Command com[maxn]; int p[maxn]; int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); } struct Node{ Node *ch[2]; int r, v, s; Node(int v) : v(v){ ch[0] = ch[1] = nullptr; r = rand(); s = 1; } bool operator < (const Node &p) const{ return r < p.r; } int cmp(int x) const{ if(x == v) return -1; return x < v ? 0 : 1; } void maintain(){ s = 1; if(ch[0] != nullptr) s += ch[0]->s; if(ch[1] != nullptr) s += ch[1]->s; } }; 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 == nullptr) o = new Node(x); else{ int d = (x < o->v ? 0 : 1); Insert(o->ch[d], x); if(o->ch[d]->r > o->r) 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] != nullptr && o->ch[1] != nullptr){ int d2 = (o->ch[0]->r > o->ch[1]->r ? 1 : 0); Rotate(o, d2); Remove(o->ch[d2], x); } else{ if(o->ch[0] == nullptr) o = o->ch[1]; else o = o->ch[0]; delete u; } } else Remove(o->ch[d], x); if(o != nullptr) o->maintain(); } Node* root[maxn]; void removeTree(Node* &o){ if(o->ch[0] != nullptr) removeTree(o->ch[0]); if(o->ch[1] != nullptr) removeTree(o->ch[1]); delete o; o = nullptr; } void Merge(Node* &src, Node* &des){ if(src->ch[0] != nullptr) Merge(src->ch[0], des); if(src->ch[1] != nullptr) Merge(src->ch[1], des); Insert(des, src->v); delete src; src = http://www.mamicode.com/nullptr;"%d %d", &n, &m) == 2 && m+n){ for(int i = 1; i <= n; ++i) scanf("%d", w+i); for(int i = 1; i <= m; ++i) scanf("%d %d", a+i, b+i); memset(removed, 0, sizeof removed); int c = 0; while(1){ int x, p = 0, v = 0; char type; scanf(" %c", &type); if(type == ‘E‘) break; scanf("%d", &x); if(type == ‘D‘) removed[x] = 1; else if(type == ‘Q‘) scanf("%d", &p); else { scanf("%d", &v); p = w[x]; w[x] = v; } com[c++] = (Command){type, x, p}; } for(int i = 1; i <= n; ++i){ p[i] = i; root[i] = new Node(w[i]); } for(int i = 1; i <= m; ++i) if(!removed[i]) add(i); cnt = tot = 0; for(int i = c-1; i >= 0; --i){ if(com[i].type == ‘D‘) add(com[i].x); else if(com[i].type == ‘C‘) change(com[i].x, com[i].p); else query(com[i].x, com[i].p); } printf("Case %d: %f\n", ++kase, cnt == 0 ? 0 : (double)tot / cnt); for(int i = 1; i <= n; ++i) if(root[i] != nullptr) removeTree(root[i]); } return 0; }
UVaLive 5031 Graph and Queries (Treap)
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