2794: [Poi2012]Cloakroom

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2794: [Poi2012]Cloakroom

Time Limit: 20 Sec Memory Limit: 128 MB
Submit: 231 Solved: 155
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Description

有n件物品，每件物品有三个属性a[i], b[i], c[i] (a[i]<b[i])。
再给出q个询问，每个询问由非负整数m, k, s组成，问是否能够选出某些物品使得：
1. 对于每个选的物品i，满足a[i]<=m且b[i]>m+s。
2. 所有选出物品的c[i]的和正好是k。

Input

第一行一个正整数n (n<=1,000)，接下来n行每行三个正整数，分别表示c[i], a[i], b[i] (c[i]<=1,000, 1<=a[i]<b[i]<=10^9)。
下面一行一个正整数q (q<=1,000,000)，接下来q行每行三个非负整数m, k, s (1<=m<=10^9, 1<=k<=100,000, 0<=s<=10^9)。

Output

输出q行，每行为TAK (yes)或NIE (no)，第i行对应第i此询问的答案。

Sample Input

5

6 2 7

5 4 9

1 2 4

2 5 8

1 3 9

5

2 7 1

2 7 2

3 2 0

5 7 2

4 1 5

Sample Output

TAK

NIE

TAK

TAK

NIE

HINT

Source

鸣谢Oimaster

[Submit][Status][Discuss]
将询问看作，考虑所有a <= m的物品，用他们凑成c的和为k的情况下，最小的b的最大值是多少如果这个值大于m + s，说明有解，否则无解那么，将询问按照m升序，物品按照a升序排好，剩下的就是普通的背包问题了

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E3 + 10;const int maxm = 1E6 + 10;const int N = 100001;struct data{int a,b,c; data(){}data(int a,int b,int c): a(a),b(b),c(c){}bool operator < (const data &B) const {return a < B.a;}}D[maxn];struct Query{int m,k,s,Num; Query(){}Query(int m,int k,int s,int Num): m(m),k(k),s(s),Num(Num){}bool operator < (const Query &B) const {return m < B.m;}}Q[maxm];int n,m,Max,f[N];bool Ans[maxm];void Insert(int g){for (int i = Max; i >= D[g].c; i--)if (f[i - D[g].c]) f[i] = max(f[i],min(f[i - D[g].c],D[g].b));}int getint(){char ch = getchar(); int ret = 0;while (ch < '0' || '9' < ch) ch = getchar();while ('0' <= ch && ch <= '9')ret = ret*10 + ch - '0',ch = getchar();return ret;}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifn = getint();for (int i = 1; i <= n; i++){int c = getint(),a,b;a = getint(); b = getint();D[i] = data(a,b,c);}sort(D + 1,D + n + 1);m = getint();for (int i = 1; i <= m; i++){int m = getint(),k,s;k = getint(); s = getint();Q[i] = Query(m,k,s,i);Max = max(Max,k);}sort(Q + 1,Q + m + 1);int now = 1; f[0] = ~0U>>1;for (int i = 1; i <= m; i++){while (now <= n && Q[i].m >= D[now].a) Insert(now++);Ans[Q[i].Num] = f[Q[i].k] > Q[i].m + Q[i].s ? 1 : 0;}for (int i = 1; i <= m; i++)if (Ans[i]) puts("TAK"); else puts("NIE");return 0;}

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