题意:给定n个物品,每个物品可以取无限次,每个物品有两种属性:价值v和颜色c
现在有q个询问,每次询问是否能取出价值和为S的方案,如有多解输出不同颜色种数的最大值
题意:看到BZOJ评论区有好心人说CC上有上一题的加强版就写了一下
首先按颜色分组,每组中取或不取只有0/1
对于每组内部就是一个同余最短路
设dp[i][j][k]为当前组不取(0)/取(1),当前共有j种不同的,模意义下和为k的最小总价值
每次建图跑SPFA转移,每次询问时暴力从大到小枚举颜色种数判断是否有解
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef pair<ll,ll>P;
#define N 200010
#define M 6000010
//#define INF 1e9
#define fi first
#define se second
#define MP make_pair
#define pb push_back
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1
#define fors(i) for(auto i:e[x]) if(i!=p) const int MOD=1e8+,inv2=(MOD+)/;
int p=1e4+;
double eps=1e-;
int dx[]={-,,,};
int dy[]={,,-,}; struct arr
{
int x,y;
}a[N]; bool cmp(arr a,arr b)
{
return a.y<b.y;
} struct node
{
int x,y,z;
node(){}
node(int a,int b,int c){x=a,y=b,z=c;}
}q[]; ll dis[][][N],INF;
int vis[][][N],mn,c; int read()
{
int v=,f=;
char c=getchar();
while(c<||<c) {if(c=='-') f=-; c=getchar();}
while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
return v*f;
} ll readll()
{
ll v=,f=;
char c=getchar();
while(c<||<c) {if(c=='-') f=-; c=getchar();}
while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
return v*f;
} void spfa(int now)
{
int t=,w=;
q[]=node(,,);
vis[][][]=;
rep(i,,)
rep(j,,c)
rep(k,,mn-)
if(dis[i][j][k]!=INF)
{
w++;
q[w]=node(i,j,k);
vis[i][j][k]=;
}
while(t<w)
{
t++;
int x=q[t].x,y=q[t].y,z=q[t].z;
vis[x][y][z]=;
int A=,B=y+-x,C=(z+now)%mn;
if(dis[x][y][z]+now<dis[A][B][C])
{
dis[A][B][C]=dis[x][y][z]+now;
if(!vis[A][B][C])
{
w++;
q[w]=node(A,B,C);
vis[A][B][C]=;
}
} }
} int main()
{
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
int n=read();
mn=1e9;
rep(i,,n) a[i].x=read(),a[i].y=read(),mn=min(mn,a[i].x);
sort(a+,a+n+,cmp);
mem(dis,0x3f);
INF=dis[][][];
c=;
dis[][][]=;
rep(i,,n-)
{
if(a[i].y!=a[i+].y)
{
rep(j,,c)
rep(k,,mn-) dis[][j][k]=min(dis[][j][k],dis[][j][k]);
}
spfa(a[i+].x);
if(a[i].y!=a[i+].y) c++;
}
rep(j,,c)
rep(k,,mn-) dis[][j][k]=min(dis[][j][k],dis[][j][k]);
int Q=read();
while(Q--)
{
ll x=readll();
int u=x%mn;
int flag=;
per(i,n,)
{
if(dis[][i][u]<=x)
{
printf("%d\n",i);
flag=;
break;
}
}
if(!flag) printf("-1\n");
}
return ;
}