POJ 3040 Allowance

一道简(kun)单(nan)的贪心题目

大体思路是从大到小枚举

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read(){
    int x=0,f=1,ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct stu{
    int v,c;
};
inline int cmp(stu a,stu b){
    return a.v<b.v;
}
stu a[25];
int n,c;
int need[25];
inline long long cal(){
    memset(need,0,sizeof(need));
    if(a[n].v>=c) return a[n--].c;
    int k=c;
    for(int i=n;i>=1;i--){
        // cout<<a[i].v<<endl;
        need[i]=min(a[i].c,k/a[i].v);
        k-=need[i]*a[i].v;
    }
    // puts("X");
    if(k){
        for(int i=1;i<=n;i++)
            if(need[i]<a[i].c){
                k=0;
                need[i]++;
                break;
            }
    }
    if(k){
        n=0;
        return 0;
    }
    int res=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
        if(need[i])
            res=min(res,a[i].c/need[i]);
    for(int i=1;i<=n;i++) a[i].c-=need[i]*res;
    // cout<<n<<"\t"<<a[n].c<<endl;
    while(n&&!a[n].c) n--;
    return res;
}
int main(){
    n=read(),c=read();
    for(int i=1;i<=n;i++) a[i].v=read(),a[i].c=read();
    sort(a+1,a+n+1,cmp);
    long long ans=0;
    while(n) ans+=cal();
    printf("%lld\n",ans);
    return 0;
}