2017 ACM Arabella Collegiate Programming Contest F. Monkeying Around GYM101350F

最新推荐文章于 2019-04-06 21:32:00 发布

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本文介绍了一种解决区间覆盖问题的方法，通过跟踪每个点最后一次受到的操作类型，并判断这些点是否被重复覆盖两次及以上。利用并查集进行区间合并，再统计各区间内被覆盖超过两次的点数。

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对单个点考虑
实际上每个点的结果只被最后的那种操作影响

那么处理一下每个点对应的最后一个操作种类
然后对每个种类都和对应的点都检查一下是否被操作了2次

转化成n个点 m个区间多少点被覆盖>=2次的问题

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.10f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
const ll infl = 10000000000000000;
const int maxn=  100000+10;
const int maxm = 40000+10;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    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;
}

vector<pii>joke[maxn];
int fa[maxn];
bool vis[maxn];
int faf(int x)
{
    return fa[x]==x?x:fa[x] = faf(fa[x]);
}
struct node
{
    int idx,l,r;
}no[maxn];
void un(int a,int b)
{
    int f1 = faf(a), f2 = faf(b);
    if(f1!=f2)fa[f2]=f1;
}
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int t;
    sd(t);
    while(t--)
    {
        int n,m;
        sdd(n,m);
        r1(i,n)fa[i] = i,vis[i] = 0;
        int mx = 0;
        r1(i,m)
        {
            int x,l,k;
            sddd(x,l,k);
            mx = max(mx,l);
            joke[l].pb(mp(max(x-k,1),-1));
            joke[l].pb(mp(min(x+k,n),1));
            no[i] = (node){l,max(x-k,1),min(x+k,n)};
        }
        int cnt = n;
        rn(i,m)
        {
            int l = no[i].l ,r = no[i].r , x = no[i].idx;
            int now = l;
            while(now<=r)
            {
                if(vis[now])now = faf(now)+1;
                else
                {
                    vis[now] = 1,--cnt;
                    joke[x].pb(mp(now,0));
                    un(r,now);
                    ++now;
                }
            }
            if(!cnt)break;
        }
        int ans = cnt;
        r1(i,mx)
        {
            sort(all(joke[i]));
            int now = 0;
            int add = 0;
            rsz(j,joke[i])
            {
                pii p = joke[i][j];
                if(p.Y==0&&now>=2)++add;
                else now -= p.Y;
            }
            ans += add;
            joke[i].clear();
        }
        ansn();
    }

    return 0;
}