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ZOJ-3721 Final Exam Arrangement 贪心

2013-07-23 00:31 274 查看
  题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3721

  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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