ZOJ-3721 Final Exam Arrangement 贪心
2013-07-23 00:31
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题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3721
容易的贪心题,排个序。。
容易的贪心题,排个序。。
//STATUS:C++_AC_840MS_6272KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cstring> #include <cassert> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <list> #include <set> #include <map> using namespace std; //using namespace __gnu_cxx; //define #define pii pair<int,int> #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define PI acos(-1.0) //typedef //typedef __int64 LL; //typedef unsigned __int64 ULL; //const const int N=100010; const int INF=0x3f3f3f3f; const int MOD=100000,STA=8000010; //const LL LNF=1LL<<60; const double EPS=1e-8; const double OO=1e15; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //Daily Use ... inline int sign(double x){return (x>EPS)-(x<-EPS);} template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} template<class T> inline T lcm(T a,T b,T d){return a/d*b;} template<class T> inline T Min(T a,T b){return a<b?a:b;} template<class T> inline T Max(T a,T b){return a>b?a:b;} 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));} //End struct Node{ int l,r; int idx; bool operator < (const Node &a)const { return l!=a.l?l<a.l:r<a.r; } }p ; int cnt ; vector<int> q ; int n; int main() { // freopen("in.txt","r",stdin); int i,j,tot,R; while(~scanf("%d",&n)) { tot=0; mem(cnt,0); for(i=0;i<=n;i++)q[i].clear(); for(i=0;i<n;i++){ scanf("%d%d",&p[i].l,&p[i].r); p[i].idx=i+1; } sort(p,p+n); R=-1; for(i=0;i<n;i++){ if(p[i].l>=R){ tot++; R=p[i].r; } R=Min(R,p[i].r); q[tot].push_back(p[i].idx); cnt[tot]++; } printf("%d\n",tot); for(i=1;i<=tot;i++){ printf("%d",q[i][0]); for(j=1;j<cnt[i];j++) printf(" %d",q[i][j]); putchar('\n'); } putchar('\n'); } return 0; }
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