sicily 1663. Party Party Party

#include<iostream>        
#include<stdio.h>
#include<cstring>
using namespace std;
int party[100][2],vis[100],p,s,e;
int main()
{
    int t=1,i,j,k;
    while(cin>>p,p)
    {
        for(i=0;i<p;++i)
        {
            cin>>s>>e;
            party[i][0]=8+(s-8)*2;     //把一刻钟拆成两半
            party[i][1]=8+(e-8)*2;
        }
        memset(vis,0,sizeof(vis));
        int res=0;
        for(i=8;i<40;++i)
        {
            int end=100;
            for(j=0;j<p;++j)
                if(vis[j]==0&&party[j][0]<=i&&party[j][1]>i&&party[j][1]<end)    //在所有未参加的party中优先选择覆盖 i 这一时刻且最早结束
                {
                    end=party[j][1];
                    k=j;
                }
            if(end<100)
            {
                vis[k]=1;
                res++;
            }
        }
        printf("On day %d Emma can attend as many as %d parties.\n",t++,res); 
    }
    return 0;
}

　　

#include<iostream>        //WA
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node
{
    int st,ed;
    bool operator<(const Node& other)const
    {
        if(st==other.st)
            return ed<other.ed;        
        else
            return st<other.st;
    }
}node[200];
int vis[50];
int main()
{
    int t=1,p,s,e;
    while(cin>>p,p)
    {
        for(int i=0;i<p;++i)
        {
            cin>>s>>e;
            node[i].st=8+(s-8)*2;
            node[i].ed=8+(e-8)*2;
        }
        sort(node,node+p);
        memset(vis,0,sizeof(vis));
        int res=0;
        for(int i=0;i<p;++i)
        {
            for(int j=node[i].st;j<node[i].ed;++j)
                if(vis[j]==0)
                {
                    vis[j]=1;
                    res++;
                    break;
                }
        }
        printf("On day %d Emma can attend as many as %d parties.\n",t++,res); 
    }
    return 0;
}

/*
选择策略有问题,比如输入:

5
8 11
8 11
8 11
8 11
9 10

答案应该是5而不是4

*/