http://47.95.147.191/contest/4/problem/A

http://47.95.147.191/contest/4/problem/A
f[i][j]表示已经完成j道且最后一个月完成[i,j]题目的最少月数。

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)

using namespace std;
long long n,k,ans;
long long f[1010][1010],sum1[1010],sum2[1010];//已经完成i-1道,最后一个月完成[i,j]道的最小天数
struct node{
    long long x;
    long long y;
}a[N];
void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

signed main(){
    in(k);in(n);
    For(i,1,n){
        in(a[i].x);in(a[i].y);
        sum1[i]=sum1[i-1]+a[i].x;
        sum2[i]=sum2[i-1]+a[i].y;
    }
    For(i,1,n)
        For(j,1,n)
            f[i][j]=inf;
    For(i,1,n) if(sum1[i]<=k&&sum2[i]<=k) f[1][i]=3;
    ans=inf;
    //[i,j]
    For(j,2,n)
        For(i,2,j)
            For(t,1,i-1){
                if(sum1[j]-sum1[i-1]<=k&&sum2[j]-sum2[i-1]<=k) f[i][j]=min(f[i][j],f[t][i-1]+2);
                if(sum1[j]-sum1[i-1]+sum2[i-1]-sum2[t-1]<=k&&sum2[j]-sum2[i-1]<=k) f[i][j]=min(f[i][j],f[t][i-1]+1);
                if(j==n) ans=min(ans,f[i][j]);
            }
    o(ans);
    return 0;
}

 

posted @ 2020-02-25 22:15  WeiAR  阅读(192)  评论(0编辑  收藏  举报