ZOJ

https://vjudge.net/problem/ZOJ-3953

题意:给了n个区间，要求你删去最少的区间，使任意三个区间 a,b,c 不存在 a与b相交，b与c相交，c与a相交 的情况

做法：xgg太**猛了呀，刚开始就觉得这个是贪心，但我肯定写不出来（事实也如此，感觉现在想通好多了。    看到l[i] r[i]范围，离散化嘛，这个没得说，一分析一画图，如果在

同等左端点的情况下，删右端点更远的肯定最优（如果要删的话），因为越往右边，对后面的影响更大，或者换句话说，对后面产生影响的可能性更大。所以删右端点远的边，

按照这种思想，我们可以得出贪心的结论，然后就是中间的处理了，记录id，xgg选择了用优先队列处理，对>重载一下，找出r大的在top上。

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma GCC optimize(2)#pragma comment(linker, "/STACK:102400000,102400000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=5e4+10;const int maxx=1e3+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(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;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;struct node {int l,r,id;friend bool operator <(node a1,node b1){return a1.r<b1.r;}}q[maxn];bool cmp(node a,node b){if(a.l==b.l) return a.r>b.r;return a.l<b.l;}int l[maxn],r[maxn];int d[maxn*4],b[maxn*4];int v[maxn];int n;void solve(){me(d,0);s_1(n);int cnt=0;FOR(1,n,i){s_2(l[i],r[i]);b[++cnt]=l[i]-1;b[++cnt]=l[i];b[++cnt]=r[i]-1;b[++cnt]=r[i];}sort(b+1,b+cnt+1);int pos=unique(b+1,b+cnt+1)-(b+1);FOR(1,n,i){q[i].l=lower_bound(b+1,b+pos+1,l[i])-(b+1);q[i].r=lower_bound(b+1,b+pos+1,r[i])-(b+1);q[i].id=i;}int num=0;int ans=0;priority_queue<node>Q;sort(q+1,q+n+1,cmp);cnt=1;FOR(1,pos,i){W(cnt<=n&&q[cnt].l<=i){Q.push(q[cnt]);d[q[cnt].l]++;d[q[cnt].r+1]--;cnt++;}num+=d[i];W(num>=3){node tmp=Q.top();Q.pop();num--;d[tmp.r+1]++;v[++ans]=tmp.id;}}sort(v+1,v+ans+1);print(ans);if(ans==0) puts("");else FOR(1,ans,i) printf("%d%c",v[i],i==ans?'\n':' ');}int main(){   // freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    //init();    s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case #%d: ",cas);        solve();    }}