标签:puts blog 题目 cas names rem type pre set
题目来源:HDU 3726 Graph and Queries
题意:见白书
思路:刚学treap 參考白皮书
#include <cstdio> #include <cstring> #include <cstdlib> using namespace std; 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& rhs) const{ return r < rhs.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; } }; 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(); //printf("--------+%d\n", o->s); } 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(); } const int maxn = 20010; const int maxc = 500010; int n, m, w[maxn], from[maxn*3], to[maxn*3], removed[maxn*3]; int f[maxn]; Node* root[maxn]; struct Command { char type; int x, p; Command(){} Command(char type, int x, int p): type(type), x(x), p(p) { } }commands[maxc]; int find(int x) { if(x != f[x]) return f[x] = find(f[x]); return f[x]; } 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) return kth(o->ch[1], k); else return kth(o->ch[0], k-s-1); } 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 add_edge(int x){ int u = find(from[x]), v = find(to[x]); //printf("-/++++%d %d\n", u, v); if(u != v){ if(root[u]->s > root[v]->s){ f[v] = u; //printf("----%d %d\n", root[v]->s, root[u]->s); mergeto(root[v], root[u]); //printf("----%d\n", root[u]->s); } else{ f[u] = v; //printf("----%d %d\n", root[u]->s, root[v]->s); mergeto(root[u], root[v]); //printf("++++%d\n", root[v]->s); } } } 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; } int query_cnt; long long query_tol; void query(int x, int k){ query_cnt++; //printf("%d %d\n", root[find(x)]->s, kth(root[find(x)], k)); query_tol += kth(root[find(x)], k); } void change_weight(int x, int v){ int u = find(x); remove(root[u], w[x]); insert(root[u], v); w[x] = v; } int main() { int kase = 0; while(scanf("%d %d", &n, &m) && (n||m)) { for(int i = 1; i <= n; i++) scanf("%d", &w[i]); for(int i = 1; i <= m; i++) scanf("%d %d", &from[i], &to[i]); memset(removed, 0, sizeof(removed)); int c = 0; while(1) { char type[2]; int x, p = 0, v = 0; scanf("%s", type); if(type[0] == 'E') break; scanf("%d", &x); if(type[0] == 'D') removed[x] = 1; if(type[0] == 'Q') scanf("%d", &p); if(type[0] == 'C') { scanf("%d", &v); p = w[x]; w[x] = v; } commands[c++] = Command(type[0], x, p); } for(int i = 1; i <= n; i++) { f[i] = i; if(root[i] != NULL) removetree(root[i]); root[i] = new Node(w[i]); //printf("******%d\n", root[i]->s); } for(int i = 1; i <= m; i++) if(!removed[i]) { add_edge(i); //puts("gh"); } query_tol = query_cnt = 0; for(int i = c-1; i >= 0; i--){ if(commands[i].type == 'D') add_edge(commands[i].x); if(commands[i].type == 'Q') query(commands[i].x, commands[i].p); if(commands[i].type == 'C') change_weight(commands[i].x, commands[i].p); } printf("Case %d: %.6lf\n", ++kase, query_tol/(double)query_cnt); } return 0; }
HDU 3726 Graph and Queries treap树
标签:puts blog 题目 cas names rem type pre set
原文地址:http://www.cnblogs.com/mfmdaoyou/p/7287250.html