hdu5240——Exam(贪心)

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1922    Accepted Submission(s): 941

Problem Description

As this term is going to end, DRD needs to prepare for his final exams.

DRD has n exams. They are all hard, but their difficulties are different. DRD will spend at leastri hours on the i-th course before its exam starts, or he will fail it. The i-th course's exam will take place ei hours later from now, and it will last for li hours. When DRD takes an exam, he must devote himself to this exam and cannot (p)review any courses. Note that DRD can review for discontinuous time.

So he wonder whether he can pass all of his courses.

No two exams will collide.

 

Input

First line: an positive integer T≤20 indicating the number of test cases.
There are T cases following. In each case, the first line contains an positive integern≤105, and n lines follow. In each of these lines, there are 3 integers ri,ei,li, where 0≤ri,ei,li≤109.

 

Output

For each test case: output ''Case #x: ans'' (without quotes), wherex is the number of test cases, and ans is ''YES'' (without quotes) if DRD can pass all the courses, and otherwise ''NO'' (without quotes).

 

Sample Input

233 2 25 100 27 1000 233 10 25 100 27 1000 2

 

Sample Output

Case #1: NOCase #2: YES

 

给出若干门课的通过需要复习的时间，考试开始时间和考试持续时间，求这个人能否参加和通过所有考试；

排序贪心解决(考试得一门一门的来，所以按照开始时间排序，开始时间相同就按复习时间来，即优先准备靠的近的考试)，看是否有重合的区间。有就NO，没有就YES

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0xffffffff//0x3f3f3f3f
#define mod 100
const int M=1005;
int n,m;
int cnt;
int sx,sy,sz;
int mp[1000][1000];
int pa[M*10],rankk[M];
int head[M*6],vis[M*100];
int dis[M*100];
int prime[M*1000];
bool isprime[M*1000];
int lowcost[M],closet[M];
char st1[5050],st2[5050];
//int len[M*6];
typedef pair<int ,int> ac;
int dp[55][55][55][55];
int has[1050000];
int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};
int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
void getpri()
{
    int i;
    int j;
    cnt=0;
    memset(isprime,false,sizeof(isprime));
    for(i=2; i<1005; i++)
    {
        if(!isprime[i])prime[cnt++]=i;
        for(j=0; j<cnt&&prime[j]*i<10000; j++)
        {
            isprime[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
struct node
{
    int r,e,l;
};
bool cmp(node a,node b){
    if(a.e!=b.e)
    return a.e<b.e;
    return a.r<b.r;
}
vector<int> g[100005];
char str[10005];
int bit[50];
node num[100005];
int main()
{
    int t,i,j,k;
    int cas =0 ;
    scanf("%d",&t) ;
    while(t--)
    {
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%d%d%d",&num[i].r,&num[i].e,&num[i].l);
        }
        sort(num,num+n,cmp);
        k=0;bool flag=1;
        for(i=0;i<n;i++){
            if(num[i].r+k<=num[i].e)//复习完在考试前
            {
                k=num[i].e+num[i].l;//准备下一门
            }
            else {
                flag=false;break;//捉襟见肘了
            }
        }
        if(flag)
        {
            printf("Case #%d: YES\n",++cas);
        }
        else
        {
            printf("Case #%d: NO\n",++cas);
        }
    }
    return 0;
}