BZOJ 2038 小Z的袜子(hose) 莫队算法

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2038

题意：

Description

作为一个生活散漫的人，小Z每天早上都要耗费很久从一堆五颜六色的袜子中找出一双来穿。终于有一天，小Z再也无法忍受这恼人的找袜子过程，于是他决定听天由命……
具体来说，小Z把这N只袜子从1到N编号，然后从编号L到R(L 尽管小Z并不在意两只袜子是不是完整的一双，甚至不在意两只袜子是否一左一右，他却很在意袜子的颜色，毕竟穿两只不同色的袜子会很尴尬。
你的任务便是告诉小Z，他有多大的概率抽到两只颜色相同的袜子。当然，小Z希望这个概率尽量高，所以他可能会询问多个(L,R)以方便自己选择。

Input

输入文件第一行包含两个正整数N和M。N为袜子的数量，M为小Z所提的询问的数量。接下来一行包含N个正整数Ci，其中Ci表示第i只袜子的颜色，相同的颜色用相同的数字表示。再接下来M行，每行两个正整数L，R表示一个询问。

Output

包含M行，对于每个询问在一行中输出分数A/B表示从该询问的区间[L,R]中随机抽出两只袜子颜色相同的概率。若该概率为0则输出0/1，否则输出的A/B必须为最简分数。（详见样例）

Sample Input

6 4
1 2 3 3 3 2
2 6
1 3
3 5
1 6

Sample Output

2/5
0/1
1/1
4/15

思路：学习莫队算法的第一题，感觉莫队算法是一个姿势优美的暴力程序~

直接分块：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;const int N = 50010;struct node{    int l, r, id;}g[N];int n, m, unit;int a[N], b[N], num[N];ll tmp, ax[N], bx[N];bool cmp(node a, node b){    return a.l/unit != b.l/unit ? a.l/unit < b.l/unit :a.r < b.r;}void add(int i){    tmp -= (ll)num[i] * num[i];    num[i]++;    tmp += (ll)num[i] * num[i];}void del(int i){    tmp -= (ll)num[i] * num[i];    num[i]--;    tmp += (ll)num[i] * num[i];}ll gcd(ll a, ll b){    if(b == 0) return a;    else return gcd(b, a % b);}void solve(){    unit = (int)sqrt(n);    sort(g+1, g+1+m, cmp);    memset(num, 0, sizeof num);    int l = 1, r = 0;    tmp = 0;    for(int i = 1; i <= m; i++)    {        while(r < g[i].r) add(a[++r]);        while(r > g[i].r) del(a[r--]);        while(l < g[i].l) del(a[l++]);        while(l > g[i].l) add(a[--l]);        ll tmp1 = tmp - (r - l + 1), tmp2 = (ll)(r - l + 1) * (r - l);        ll ans = gcd(tmp1, tmp2);        ax[g[i].id] = tmp1 / ans, bx[g[i].id] = tmp2 / ans;    }    for(int i = 1; i <= m; i++) printf("%lld/%lld\n", ax[i], bx[i]);}int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];    sort(b+1, b+1+n);    for(int i = 1; i <= n; i++) a[i] = lower_bound(b+1, b+1+n, a[i]) - b;    for(int i = 1; i <= m; i++) scanf("%d%d", &g[i].l, &g[i].r), g[i].id = i;    solve();    return 0;}

以曼哈顿距离最小生成树为基础的：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>using namespace std;typedef long long ll;const int N = 50010;const int INF = 0x3f3f3f3f;struct node{    int x, y, id;}p[N], pp[N];struct edge{    int v, u, d;}g[N*10];struct graph{    int to, next;}gra[N*4];struct BIT{    int minn, pos;    void init()    {        minn = INF, pos = -1;    }}bit[N];int n, m, a[N], tot;int par[N];int cnt, head[N];int num[N];ll tmp;bool cmp(node a, node b){    if(a.x != b.x) return a.x < b.x;    else return a.y < b.y;}bool cmpg(edge a, edge b){    return a.d < b.d;}int ser(int x){    int r = x, i = x, j;    while(r != par[r]) r = par[r];    while(par[i] != r) j = par[i], par[i] = r, i = j;    return r;}int ask(int i, int m){    int minn = INF, pos = -1;    while(i <= m)    {        if(bit[i].minn < minn)            minn = bit[i].minn, pos = bit[i].pos;        i += i & -i;    }    return pos;}void update(int i, int val, int pos){    while(i > 0)    {        if(val < bit[i].minn)            bit[i].minn = val, bit[i].pos = pos;        i -= i & -i;    }}void add_edge(int v, int u, int d){    g[tot].v = v, g[tot].u = u, g[tot++].d = d;}void _add_edge(int v, int u){    gra[cnt].to = u, gra[cnt].next = head[v], head[v] = cnt++;}void mmst(int n, node p[]){    int a[N], b[N];    tot = 0;    for(int i = 0; i < 4; i++)    {        if(i == 1 || i == 3)        {            for(int j = 0; j < n; j++)                swap(p[j].x, p[j].y);        }        else if(i == 2)        {            for(int j = 0; j < n; j++)                p[j].x = -p[j].x;        }        sort(p, p + n, cmp);        for(int j = 0; j < n; j++)            a[j] = b[j] = p[j].y - p[j].x;        sort(b, b + n);        int m = unique(b, b + n) - b;        for(int j = 1; j <= m; j++)            bit[j].init();        for(int j = n - 1; j >= 0; j--)        {            int pos = lower_bound(b, b + m, a[j]) - b + 1;            int ans = ask(pos, m);            if(ans != -1)                add_edge(p[j].id, p[ans].id, abs(p[j].x - p[ans].x) + abs(p[j].y - p[ans].y));            update(pos, p[j].x + p[j].y, j);        }    }    cnt = 0;    memset(head, -1, sizeof head);    for(int i = 0; i < n; i++) par[i] = i;    sort(g, g + tot, cmpg);    for(int i = 0; i < tot; i++)    {        int fv = ser(g[i].v), fu = ser(g[i].u);        if(fv != fu)        {            par[fv] = fu;            _add_edge(g[i].v, g[i].u);            _add_edge(g[i].u, g[i].v);        }    }}struct Ans{    ll a, b;}ans[N];void cal(int l, int r, int d){    for(int i = l; i <= r; i++)    {        tmp -= (ll)num[a[i]] * num[a[i]];        num[a[i]] += d;        tmp += (ll)num[a[i]] * num[a[i]];    }}void dfs(int l1, int r1, int l2, int r2, int idx, int pre){    if(l2 < l1) cal(l2, l1 - 1, 1);    if(r2 > r1) cal(r1 + 1, r2, 1);    if(l2 > l1) cal(l1, l2 - 1, -1);    if(r2 < r1) cal(r2 + 1, r1, -1);    ans[pp[idx].id].a = tmp - (r2 - l2 + 1);    ans[pp[idx].id].b = (ll)(r2 - l2 + 1) * (r2 - l2);    for(int i = head[idx]; i != -1; i = gra[i].next)    {        int u = gra[i].to;        if(pre == u) continue;        dfs(l2, r2, pp[u].x, pp[u].y, u, idx);    }    if(l2 < l1) cal(l2, l1 - 1, -1);    if(r2 > r1) cal(r1 + 1, r2, -1);    if(l2 > l1) cal(l1, l2 - 1, 1);    if(r2 < r1) cal(r2 + 1, r1, 1);}ll gcd(ll a, ll b){    if(b == 0) return a;    else return gcd(b, a % b);}int main(){    while(~ scanf("%d%d", &n, &m))    {        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);        for(int i = 0; i < m; i++)        {            scanf("%d%d", &p[i].x, &p[i].y);            p[i].id = i;            pp[i] = p[i];        }        mmst(m, p);        memset(num, 0, sizeof num);        tmp = 0;        dfs(1, 0, pp[0].x, pp[0].y, 0, -1);        for(int i = 0; i < m; i++)        {            ll d = gcd(ans[i].a, ans[i].b);            printf("%lld/%lld\n", ans[i].a / d, ans[i].b / d);        }    }    return 0;}