hdu-2419 Boring Game

http://acm.hdu.edu.cn/showproblem.php?pid=2419

给一个图，预分配点值。随后有三种操作，F u v查询与u联通部分大于等于v的最小的数，没有则返回0，U u v更新u的值为v，E u v删除u-v的边。

联通块可以用并查集解决，但是删边无法处理。因为没有加边，我们可以把整个操作过程反过来进行，就变成只有加边没有删边了。期间仔细维护各个值就好。

涉及边的删除，所以用set存边。查找最值可以用multi_set的lower_bound。并查集涉及集合合并，可以用s[a].insert(b.begin(),b.end())，实测效率和遍历b并Insert到a中几乎没差别，切记要让小集合向大集合合并！！！

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define LL int
using namespace std;
const LL N = 20004;
multiset<int> g[N];
int n, m, q;
LL val[N];
struct cmd
{
    char e;
    LL k, v;
    cmd(char ee, LL kk, LL vv)
    {
        e = ee;
        k = kk;
        v = vv;
    }
    cmd()
    {

    }
};
LL pre[N];
multiset<LL> s[N];
LL Find(LL x)
{
    if (pre[x] == x)return x;
    return pre[x]=Find(pre[x]);
}
void mix(LL t1, LL t2)
{
    LL pt1 = Find(t1), pt2 = Find(t2);
    if (pt1==pt2)return;
    if (s[pt1].size() > s[pt2].size()) swap(pt1, pt2);
    pre[pt1] = pt2;
    s[pt2].insert(s[pt1].begin(), s[pt1].end());
    //for (multiset<LL>::iterator it = s[pt1].begin(); it != s[pt1].end(); it++)
        //s[pt2].insert(*it);
    s[pt1].clear();
}
void upt(LL nod, LL v)
{
    LL pt = Find(nod);
    //cout << pt << endl;
    s[pt].erase(s[pt].find(val[nod]));
    val[nod] = v;
    s[pt].insert(v);
}
LL fd(LL nod, LL v)
{
    LL pt = Find(nod);
    multiset<LL>::iterator it= s[pt].lower_bound(v);
    if (it == s[pt].end())return 0;
    return *it;
}
cmd v[N * 30];
int main() {
    int cas = 1;
    while (scanf("%d%d%d",&n,&m,&q)!=EOF)
    {
        for (int i = 1; i <= n; i++)
        {
            LL num;
            s[i].clear();
            g[i].clear();
            scanf("%d", &num);
            val[i] = num;
            pre[i] = i;
        }
        for (int i = 1; i <= m; i++)
        {
            int fr, to;
            scanf("%d%d", &fr, &to);
            g[min(fr,to)].insert(max(fr,to));
            //g[to].insert(fr);
        }
        LL siz = 0;
        while (q--)
        {
            char c[5];
            LL t1, t2;
            scanf("%s %d %d", c, &t1, &t2);
            if (c[0] == 'E')
            {
                //g[t1].erase(t2);
                g[min(t1,t2)].erase(g[min(t1,t2)].find(max(t1,t2)));
            }
            if (c[0] == 'U')
            {
                LL temp = val[t1];
                val[t1] = t2;
                t2 = temp;
            }
            v[siz++]=cmd(c[0], t1, t2);
        }
        for (int i = 1; i <= n; i++)s[i].insert(val[i]);
        for (int i = 1; i <= n; i++)
        {
            for (multiset<int>::iterator j = g[i].begin(); j != g[i].end(); j++)
            {
                int e = *j;
                mix(i, e);
            }
        }
        LL cnt = 0;
        double ans = 0;
        for (int i = siz-1; i >=0; i--)
        {
            cmd e = v[i];
            //cout << e.e << ": " << e.k <<' '<< e.v << endl;
            if (e.e == 'E')
            {
                mix(e.k, e.v);
            }
            if (e.e == 'U')
            {
                upt(e.k, e.v);
            }
            if (e.e == 'F')
            {
                ans += fd(e.k, e.v);
                //cout << fd(e.k, e.v) << endl;
                cnt++;
            }
        }
        printf("Case %d: %.3f\n", cas++,ans*1.0 / cnt);
    }

    return 0;
}