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题目链接：点击打开链接
先把询问处理成曼哈顿最小生成树。
然后在树上暴力跑即可。
能使用莫队的情况应该是对于询问[l,r] -> [l‘, r‘] 花费必须是 abs(l-l‘) + abs(r-r‘)
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) { putchar('-'); x = -x; }
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int N = 1e5 + 10;
typedef long long ll;
ll gcd(ll x, ll y){
if (x > y)swap(x, y);
while (x)y %= x, swap(x, y);
return y;
}
vector<int>G[N];
class MST{
struct Edge{
int from, to, dis;
Edge(int _from = 0, int _to = 0, int _dis = 0) :from(_from), to(_to), dis(_dis){}
bool operator < (const Edge &x) const{return dis < x.dis;}
}edge[N << 3];
int f[N], tot;
int find(int x){ return x == f[x] ? x : f[x] = find(f[x]); }
bool Union(int x, int y){
x = find(x); y = find(y);
if (x == y)return false;
if (x > y)swap(x, y);
f[x] = y;
return true;
}
public:
void init(int n){
for (int i = 0; i <= n; i++)f[i] = i;
tot = 0;
}
void add(int u, int v, int dis){
edge[tot++] = Edge(u, v, dis);
}
ll work(){//计算最小生成树，返回花费
sort(edge, edge + tot);
ll cost = 0;
for (int i = 0; i < tot; i++)
if (Union(edge[i].from, edge[i].to)){
cost += edge[i].dis;
G[edge[i].from].push_back(edge[i].to);
G[edge[i].to].push_back(edge[i].from);
}
return cost;
}
}mst;
struct Point{//二维平面的点
int x, y, id;
bool operator < (const Point&a) const{
return x == a.x ? y < a.y : x < a.x;
}
}p[N];
bool cmp_id(const Point&a, const Point&b){
return a.id < b.id;
}
class BIT{//树状数组
int c[N], id[N], maxn;
int lowbit(int x){ return x&-x; }
public:
void init(int n){
maxn = n + 10;
fill(c, c + maxn + 1, inf);
fill(id, id + maxn + 1, -1);
}
void updata(int x, int val, int _id){
while (x){
if (val < c[x]){ c[x] = val; id[x] = _id; }
x -= lowbit(x);
}
}
int query(int x){
int val = inf, _id = -1;
while (x <= maxn){
if (val > c[x]){ val = c[x]; _id = id[x]; }
x += lowbit(x);
}
return _id;
}
}tree;
inline bool cmp(int *x, int *y){ return *x < *y; }
class Manhattan_MST{
int A[N], B[N];
public:
ll work(int l, int r){
mst.init(r);
for (int dir = 1; dir <= 4; dir++){
if (dir%2==0)for (int i = l; i <= r; i++)swap(p[i].x, p[i].y);
else if (dir == 3)for (int i = l; i <= r; i++)p[i].y = -p[i].y;
sort(p + l, p + r + 1);
for (int i = l; i <= r; i++) A[i] = B[i] = p[i].y - p[i].x; //离散化
sort(B + 1, B + N + 1);
int sz = unique(B + 1, B + N + 1) - B - 1;
//初始化反树状数组
tree.init(sz);
for (int i = r; i >= l; i--)
{
int pos = lower_bound(B + 1, B + sz + 1, A[i]) - B;
int id = tree.query(pos);
if (id != -1)
mst.add(p[i].id, p[id].id, abs(p[i].x - p[id].x) + abs(p[i].y - p[id].y));
tree.updata(pos, p[i].x + p[i].y, i);
}
}
for (int i = l; i <= r; i++)p[i].y = -p[i].y;
return mst.work();
}
}m_mst;
ll up[N], now;
int l[N], r[N];
int n, query, col[N], siz[N];
void add(int x, int y){
for (int i = x; i <= y; i++)
{
now += siz[col[i]];
siz[col[i]]++;
}
}
void del(int x, int y){
for (int i = x; i <= y; i++)
{
now -= siz[col[i]] - 1;
siz[col[i]]--;
}
}
void dfs(int u, int fa){
if (fa == -1)
add(l[u], r[u]);
else
{
if (l[u] < l[fa]) add(l[u], l[fa] - 1);
if (r[u] > r[fa]) add(r[fa] + 1, r[u]);
if (l[u] > l[fa]) del(l[fa], l[u] - 1);
if (r[u] < r[fa]) del(r[u] + 1, r[fa]);
}
up[u] = now;
for (int i = 0; i < G[u].size(); i++)
if (G[u][i] != fa)dfs(G[u][i], u);
if (fa != -1)
{
if (l[u] < l[fa]) del(l[u], l[fa] - 1);
if (r[u] > r[fa]) del(r[fa] + 1, r[u]);
if (l[u] > l[fa]) add(l[fa], l[u] - 1);
if (r[u] < r[fa]) add(r[u] + 1, r[fa]);
}
}
int main(){
while (cin >> n >> query){
for (int i = 1; i <= n; i++)rd(col[i]);
for (int i = 1; i <= query; i++){
rd(l[i]), rd(r[i]);
p[i].x = l[i]; p[i].y = r[i]; p[i].id = i;
}
for (int i = 1; i <= query; i++)G[i].clear();
m_mst.work(1, query);
now = 0;
memset(siz, 0, sizeof siz);
dfs(1, -1);
for (int i = 1; i <= query; i++){
ll down = (ll)(r[i] - l[i] + 1)*(r[i] - l[i]) / 2;
ll g = gcd(up[i], down);
pt(up[i] / g); putchar('/'); pt(down / g); putchar('\n');
}
}
return 0;
}2038: [2009国家集训队]小Z的袜子(hose) 莫队算法
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原文地址：http://blog.csdn.net/qq574857122/article/details/45726427
