hdu 5333 Undirected Graph (LCT)

题意：

n个点和m条边的无向图，q个询问，每次询问[L,R],问删去至少有一个端点不在[L,R]内的边，剩下的图构成多少个联通分量。

思路：

高中生出的题。。能离线绝逼要离线搞。。否则有在线的搞法的话就一定会强制在线。。

询问按照R排序，边按照端点较大的排序，从小到大来扫询问，比如，扫到Ri时，将较大端点小于等于Ri的边以较小端点为权值加入图中，如果形成环的话，就要删去环上最小的边，也就是维护图的最大生成森林，因为对于询问[L,R]，如果权值较大的边都小于L的话，那么权值更小的边也是不考虑的，用LCT维护最大生成森林，tot记录森林中的总边数，树状数组记录边权个数，答案就是n-(tot-t)，t是生成森林中小于L的边的个数。

#include<iostream>#include<string>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#include<set>#include<algorithm>using namespace std;#define LL long long#define eps 1e-8#define MP make_pair#define N 500020#define M 80020#pragma comment(linker, "/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)#define lson ll, md, ls#define rson md + 1, rr, rs#define mod 258280327#define inf 0x3f3f3f3f#define ULL unsigned long longint readint() {    char c;    while((c = getchar()) && !(c >= '0' && c <= '9'));    int ret = c - '0';    while((c = getchar()) && c >= '0' && c <= '9')        ret = ret * 10 + c - '0';    return ret;}struct edge {    int u, v;    edge() {}    void input() {        u = readint();        v = readint();        if(u < v) swap(u, v);    }    bool operator < (const edge &b) const {        return u < b.u;    }}a[N];struct query {    int l, r, ans, id;    void input(int id) {        this -> id = id;        l = readint();        r = readint();        ans = 0;    }    bool operator < (const query &b) const {        return r < b.r;    }}b[N];int ch[N][2], rt[N], pre[N], rev[N], mi[N];int val[N];int key[N];int n, m, q;int bit[N];void update_rev(int x) {    if(!x) return;    rev[x] ^= 1;    swap(ch[x][0], ch[x][1]);}bool judge(int x, int y) {    while(pre[x]) x = pre[x];    while(pre[y]) y = pre[y];    return x == y;}void push_up(int x) {    mi[x] = key[x];    if(val[mi[ch[x][0]]] < val[mi[x]]) mi[x] = mi[ch[x][0]];    if(val[mi[ch[x][1]]] < val[mi[x]]) mi[x] = mi[ch[x][1]];}void push_down(int x) {    if(rev[x]) {        update_rev(ch[x][0]);        update_rev(ch[x][1]);        rev[x] = 0;    }}void P(int x) {    if(!rt[x]) P(pre[x]);    push_down(x);}void rot(int x) {    int y = pre[x], d = ch[y][1] == x;    ch[y][d] = ch[x][!d];    pre[ch[x][!d]] = y;    pre[x] = pre[y];    pre[y] = x;    ch[x][!d] = y;    if(rt[y]) rt[y] = false, rt[x] = true;    else        ch[pre[x]][ch[pre[x]][1]==y] = x;    push_up(y);}void splay(int x) {    P(x);    while(!rt[x]) {        int f = pre[x], ff = pre[f];        if(rt[f]) rot(x);        else if((ch[ff][1] == f) == (ch[f][1] == x))            rot(f), rot(x);        else            rot(x), rot(x);    }    push_up(x);}void Access(int x) {    int y = 0;    for(; x; y = x, x = pre[x]) {        splay(x);        rt[ch[x][1]] = true;        ch[x][1] = y;        rt[y] = false;        push_up(x);    }}void make_root(int x) {    Access(x);    splay(x);    update_rev(x);}void link(int x, int y) {    make_root(x);    pre[x] = y;}void cut(int x, int y) {    make_root(x);    splay(y);    pre[ch[y][0]] = pre[y];    pre[y] = 0;    rt[ch[y][0]] = true;    ch[y][0] = 0;    push_up(y);}void lca(int &x, int &y) {    Access(y), y = 0;    for(splay(x); pre[x]; y = x, x = pre[x], splay(x)) {        rt[ch[x][1]] = true;        ch[x][1] = y;        rt[y] = false;        push_up(x);    }}//okint query_mi(int x, int y) {    /*    make_root(x);    Access(y);    splay(y);    return mi[y];    */    lca(x, y);    int ret = key[x];    if(val[mi[y]] < val[ret]) ret = mi[y];    if(val[mi[ch[x][1]]] < val[ret]) ret = mi[ch[x][1]];    return ret;}bool cmp(query x, query y) {    return x.id < y.id;}//okvoid bit_add(int x, int v) {   // printf("bit_add %d %d\n", x, v);    while(x <= n) {        bit[x] += v;        x += x & -x;    }}//okint bit_query(int x) {    int ret = 0;    while(x) {        ret += bit[x];        x -= x & -x;    }    return ret;}int main() {    //freopen("tt.txt", "r", stdin);    while(scanf("%d%d%d", &n, &m, &q) != EOF) {        for(int i = 1; i <= m; ++i) {            a[i].input();        }        for(int i = 1; i <= q; ++i) {            b[i].input(i);        }        sort(a + 1, a + m + 1);        sort(b + 1, b + q + 1);        for(int i = 0; i <= n; ++i) val[i] = inf;        for(int i = 1; i <= m; ++i) val[i+n] = a[i].v;        int j = 1;        for(int i = 1; i <= n + m; ++i) {            pre[i] = 0;            rt[i] = 1;            key[i] = i;            mi[i] = i;            ch[i][0] = ch[i][1] = 0;            rev[i] = 0;        }        for(int i = 1; i <= n; ++i) bit[i] = 0;        int tot = 0;        for(int i = 1; i <= q; ++i) {            while(j <= m && a[j].u <= b[i].r) {                int u = a[j].u, v = a[j].v;                if(!judge(u, v)) {                    link(u, j + n);                    link(v, j + n);                    bit_add(val[j+n], 1);                    ++tot;                }                else {                    int t = query_mi(u, v);                    if(val[t] < val[j+n]) {                        cut(a[t-n].u, t);                        cut(a[t-n].v, t);                        bit_add(val[t], -1);                        link(u, j + n);                        link(v, j + n);                        bit_add(val[j+n], 1);                    }                }                ++j;            }            int t = bit_query(b[i].l - 1);            b[i].ans = n - (tot - t);        }        sort(b + 1, b + q + 1, cmp);        for(int i = 1; i <= q; ++i)            printf("%d\n", b[i].ans);    }    return 0;}