VK Cup 2012 Qualification Round 1 E

先利用动态规划求出n个电话后能忽略k电话所能空间时间的最小左边界（哎！！当时还想用贪心，为啥就没想到动态规划呢！！）然后就是一个枚举过程，因为一个休息时间必将以一个电话的到来而结束，枚举4000个电话，然后就ok了！！！！说白了就tm是个水dp！！！太水了！！周五的vk一定会ak！！！

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif

#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define SZ(a)           ((int)a.size())
#define PP(n,m,a)		puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-10;
const double pi = acos(-1.0);
const int maxn = 4444;

int t[maxn],d[maxn];
int dp[maxn][maxn];
int n,k,ans;

int main()
{
    while(cin>>n>>k)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>t[i]>>d[i];
        }
        t[n+1]=86401;
        d[n+1]=111;
        FF(i,maxn) dp[0][i]=1;
        int qw,as;
        FOR(i,1,n)
        {
            FOR(j,0,k)
            {
                qw=dp[i-1][j];
                if(qw <= t[i])
                {
                    qw=t[i]+d[i];
                }
                else
                {
                    qw=qw+d[i];
                }
                if(j)
                {
                    as=dp[i-1][j-1];
                }
                else
                {
                    as=inf;
                }
                dp[i][j]=min(as,qw);
            }
        }
        ans=0;
        FOR(i,1,n+1)
        {
            ans=max(ans,t[i]-dp[i-1][k]);
        }
        if(n==k)
        {
            ans=86400;
        }
        printf("%d\n",ans);
    }
    return 0;
}