【线段树合并】联通分量计数

给定一棵n个节点的树，从1到n标号。选择编号连续的若干个点，你需要选择一些边使得这些点通过选择的边联通，目标是使得选择的边数最少。

现需要计算对于n*(n+1)/2种选择的情况，最小选择边数的总和为多少。

Input
第一行一个数n(1<=n<=100000)
接下来n-1行，每行两个数x,y描述一条边(1<=x,y<=n)
Output
一个数，表示答案
Input示例
3
3 1
3 2
Output示例
5

注：若有n棵含单个元素的树，经过n-1次合并，合并成一棵树，总代价为O(nlogn)或O(nlogU)
定义线段树a的势能f(a)为 线段树a的总节点数
合并两棵值域相同的线段树的时间复杂度为O(两棵线段树重合节点的个数), 即势能的减少数量。
单叶节点线段树的节点数为logU，即f = logU
故合并线段树的时间复杂度为 O(nlogU)

  1 #include <bits/stdc++.h>
  2 
  3 #define ll long long
  4 #define ull unsigned long long
  5 #define st first
  6 #define nd second
  7 #define pii pair<int, int>
  8 #define pil pair<int, ll>
  9 #define pli pair<ll, int>
 10 #define pll pair<ll, ll>
 11 #define tiii tuple<int, int, int>
 12 #define pw(x) ((1LL)<<(x))
 13 #define lson l, m, rt<<1
 14 #define rson m+1, r, rt<<1|1
 15 #define FIN freopen("a.in","r",stdin);
 16 #define FOUT freopen("a.out","w",stdout);
 17 using namespace std;
 18 /***********/
 19 template <class T>
 20 bool scan (T &ret) {
 21     char c;
 22     int sgn;
 23     if (c = getchar(), c == EOF) return 0; //EOF
 24     while (c != '-' && (c < '0' || c > '9') ) c = getchar();
 25     sgn = (c == '-') ? -1 : 1;
 26     ret = (c == '-') ? 0 : (c - '0');
 27     while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
 28     ret *= sgn;
 29     return 1;
 30 }
 31 template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
 32 template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
 33 template<typename T>
 34 inline int sgn(T a) {return a>0?1:(a<0?-1:0);}
 35 template <class T1, class T2>
 36 bool gmax(T1 &a, const T2 &b) { return a < b? a = b, 1:0;}
 37 template <class T1, class T2>
 38 bool gmin(T1 &a, const T2 &b) { return a > b? a = b, 1:0;}
 39 template <class T> inline T lowbit(T x) {return x&(-x);}
 40 
 41 template<class T1, class T2>
 42 ostream& operator <<(ostream &out, pair<T1, T2> p) {
 43     return out << "(" << p.st << ", " << p.nd << ")";
 44 }
 45 template<class A, class B, class C>
 46 ostream& operator <<(ostream &out, tuple<A, B, C> t) {
 47     return out << "(" << get<0>(t) << ", " << get<1>(t) << ", " << get<2>(t) << ")";
 48 }
 49 template<class T>
 50 ostream& operator <<(ostream &out, vector<T> vec) {
 51     out << "("; for(auto &x: vec) out << x << ", "; return out << ")";
 52 }
 53 void testTle(int &a){
 54     while(1) a = a*(ll)a%1000000007;
 55 }
 56 const int inf = 0x3f3f3f3f;
 57 const ll INF = 1e17;
 58 const ll mod = 1000000007;
 59 const double eps = 1e-6;
 60 const int N = 3e6+5;
 61 /***********/
 62 
 63 
 64 struct Node{
 65     int l, r;//子树指针
 66     ll ans;
 67     int lsize, rsize;//左侧连续长度，右侧连续长度
 68     bool lnum, rnum;//最左值
 69 };
 70 Node T[N];
 71 
 72 void debug(Node &A, int l, int r){
 73     puts("******");
 74     printf("l: %d, r: %d\n", l, r);
 75     printf("ans: %lld\n", A.ans);
 76     printf("lsize:%d, rsize:%d\n", A.lsize, A.rsize);
 77     printf("lnum:%d rnum:%d\n", A.lnum*1, A.rnum*1);
 78     puts("******");
 79     if(A.l) debug(T[A.l], l, (l+r)/2);
 80     if(A.r) debug(T[A.r], (l+r)/2+1, r);
 81 }
 82 
 83 int cnt;
 84 int n;
 85 void pushup(int rt, int l, int r){//更新rt节点信息
 86     int lp = T[rt].l, rp = T[rt].r;
 87     int m = (l+r) >> 1;
 88     //更新
 89     T[rt].ans = (lp? T[lp].ans: 0) + (rp? T[rp].ans: 0) + (m-l+1LL)*(r-m);///////
 90     if( (lp? T[lp].rnum: 0) == (rp? T[rp].lnum: 0) ){
 91         T[rt].ans -= (lp? T[lp].rsize: m-l+1LL)*(rp? T[rp].lsize: r-m);
 92         if(lp == 0||T[lp].lsize == m-l+1) T[rt].lsize = m-l+1+(rp? T[rp].lsize: r-m);
 93         else T[rt].lsize = T[lp].lsize;
 94         if(rp == 0||T[rp].rsize == r-m) T[rt].rsize = r-m+(lp? T[lp].rsize: m-l+1);
 95         else T[rt].rsize = T[rp].rsize;
 96     }
 97     else T[rt].lsize = lp? T[lp].lsize: m-l+1, T[rt].rsize = rp? T[rp].rsize: r-m;
 98     T[rt].lnum = lp? T[lp].lnum: 0, T[rt].rnum = rp? T[rp].rnum: 0;
 99 }
100 
101 int update(int x, int l, int r, int &rt){//插入一个节点
102     rt = ++cnt;
103     if(cnt == N){
104         testTle(x);
105     }
106     T[rt].l = T[rt].r = 0;
107     T[rt].ans = (x-(ll)l)*(r-x)+r-l;//x
108     T[rt].lsize = x == l? 1: x-l, T[rt].rsize = x == r? 1: r-x;
109     T[rt].lnum = x == l; T[rt].rnum = x == r;
110     if(l == r) return cnt;
111 
112     int m = (l+r) >> 1;
113     if(x <= m) update(x, l, m, T[rt].l);
114     else update(x, m+1, r, T[rt].r);
115     return cnt;
116 }
117 
118 void merg(int &x, int y, int l, int r){
119     if(!x || !y){
120         if(x == 0) x = y;
121         return ;
122     }
123     if(l == r) {testTle(x) ; return ;} //T[x].v += T[y].v; 
124     int m = (l+r) >> 1;
125     merg(T[x].l, T[y].l, l, m);
126     merg(T[x].r, T[y].r, m+1, r);
127     pushup(x, l, r);
128 }
129 
130 vector<int> ve[100005];
131 int root[100005];
132 ll ans = 0;
133 void dfs(int f, int x){
134     //add
135     update(x, 1, n, root[x]);
136     for(int y: ve[x]) if(y != f){
137         dfs(x, y);
138         merg(root[x], root[y], 1, n);
139     }
140     if(f) ans += T[ root[x] ].ans;
141 }
142 
143 int main() {
144     //FIN //FOUT
145     scanf("%d", &n);
146     for(int u, v, i = 1; i < n; i++){
147         scanf("%d%d", &u, &v);
148         ve[u].push_back(v);
149         ve[v].push_back(u);
150     }
151     dfs(0, 1);
152     printf("%lld\n", ans);
153     return 0;
154 }
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 启发式合并复杂度分析

posted @ 2017-07-29 22:16  我在地狱  阅读(303)  评论(0)  编辑  收藏  举报