“玲珑杯”郑州轻工业学院第八届ACM程序设计大赛Problem G: 蛤玮点菜

Problem G: 蛤玮点菜

Time Limit: 2 Sec Memory Limit: 128 MB
Submit: 320 Solved: 31

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Description

在我们下饭店的时候蛤玮经常负责点菜,今天饭店搞活动,当总价格大于等于X时可以减去Y的优惠,注意如果总价是2X也仅减去一倍Y.蛤玮非常了解菜品也了解大家,他知道每个菜品有一个饱食度,只有菜品饱食度的和不小于K时大家才会吃的开心.请问蛤玮如何点菜才能在让大家吃的开心的前提下花尽量少的钱,输出最后需要付的钱.注意蛤玮是个有追求的人,所以他不会点重复的菜.

Input

T(1<=T<=40),表示数据组数.

每组数据第一行n(1<=n<=30),K,X,Y(1<=K,X,Y<=1e9, X>=Y),表示一共有n种菜,X,Y,K如题目中描述.

接下来n行每行两个数ai,bi(1<=ai,bi<=1e8),分别表示第i个菜的价格和饱食度.

Output

每组数据输出一个数,表示总价.如果无解则输出"go die".

Sample Input

2 2 20 12

10 2

Sample Output

2^30的暴力会超时，只能分成两个2^15的堆，之后这两堆都按饱食度排序，价格就可以二分了

https://acm.zzuli.edu.cn/zzuliacm/problem.php?cid=1215&pid=6

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
using namespace std;
#define LL long long
#define Max 21448255551228315ll
typedef struct Res
{LL v;LL p;
}Res;
set<LL> st;
bool comp(Res a, Res b)
{if(a.v<b.v || a.v==b.v && a.p<b.p)return 1;return 0;
}
Res s1[100005], s2[100005], a[35];
int main(void)
{LL T, n, j, i, k1, k2, ed, v, p, K, X, Y, ans;scanf("%lld", &T);while(T--){ans = Max;scanf("%lld%lld%lld%lld", &n, &K, &X, &Y);for(i=1;i<=n;i++)scanf("%lld%lld", &a[i].p, &a[i].v);k1 = k2 = 0;ed = n/2;for(i=0;i<=(1ll<<ed)-1;i++){v = p = 0;for(j=0;j<=ed-1;j++){if(i&(1ll<<j)){v += a[j+1].v;p += a[j+1].p;}}s1[++k1].v = v;s1[k1].p = p;if(v>=K){if(p>=X)ans = min(ans, p-Y);elseans = min(ans, p);}}ed = n-n/2;for(i=0;i<=(1ll<<ed)-1;i++){v = p = 0;for(j=0;j<=ed-1;j++){if(i&(1ll<<j)){v += a[j+1+n/2].v;p += a[j+1+n/2].p;}}s2[++k2].v = v;s2[k2].p = p;if(v>=K){if(p>=X)ans = min(ans, p-Y);elseans = min(ans, p);}}sort(s1+1, s1+k1+1, comp);sort(s2+1, s2+k2+1, comp);st.clear();ed = k2;for(i=1;i<=k1;i++){while(s1[i].v+s2[ed].v>=K && ed>=1){p = s1[i].p+s2[ed].p;if(p>=X)p -= Y;st.insert(s2[ed].p);ans = min(ans, p);ed--;}if(st.empty()==1)continue;ans = min(ans, s1[i].p+(*st.begin()));if(st.lower_bound(X-s1[i].p)!=st.end())ans = min(ans, s1[i].p+(*st.lower_bound(X-s1[i].p))-Y);}if(ans==Max)printf("go die\n");elseprintf("%lld\n", ans);}return 0;
}