BZOJ2794: [Poi2012]Cloakroom

限制是a[i]<=m，b[i]>m+s，c[i]和为k

考虑到k<=100000，而a,b都很大，可以转化一下，变成询问是否有c[i]和为k，a[i]<=m，b[i]的最小值>m+s的情况，那么就可以按照询问的m排序，做一个DP

code:

#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;

void read(int &x)
{
char c;
while(!((c=getchar())>='0'&&c<='9'));
x=c-'0';
while((c=getchar())>='0'&&c<='9') x=(x<<3)+(x<<1)+c-'0';
}
void up(int &x,int y){if(x<y)x=y;}
void down(int &x,int y){if(x>y)x=y;}
const int maxn = 1100;
const int maxm = 1100000;

struct qu
{
int m,k,s,i;
}q[maxm]; int m;
struct node
{
int a,b,c;
}a[maxn]; int n;
bool cmp(node x,node y) { return x.a<y.a; }
bool cmpq(qu x,qu y) { return x.m<y.m; }

int v[110000];
bool ans[maxm];

int main()
{
memset(ans,true,sizeof ans);
int mx=0,mxk=0;

read(n);
for(int i=1;i<=n;i++)
{
read(a[i].c); read(a[i].a); read(a[i].b);
up(mx,a[i].b);
}sort(a+1,a+n+1,cmp);

read(m);
for(int i=1;i<=m;i++)
{
q[i].i=i;
read(q[i].m); read(q[i].k); read(q[i].s);
up(mxk,q[i].k);
if((ll)q[i].m+q[i].s>(ll)mx) ans[i]=false;
}sort(q+1,q+m+1,cmpq);

memset(v,0,sizeof v); v[0]=mx;
int nw=1;
for(int i=1;i<=n;i++)
{
while(nw<=m)
{
if(!ans[q[nw].i]) nw++;
else if(q[nw].m<a[i].a)
{
ans[q[nw].i]=v[q[nw].k]>(q[nw].m+q[nw].s);
nw++;
}
else break;
}
if(nw>m) break;
for(int j=mxk;j>=a[i].c;j--)
up(v[j],min(v[j-a[i].c],a[i].b));
}
while(nw<=m)
{
if(!ans[q[nw].i]) nw++;
else ans[q[nw].i]=v[q[nw].k]>(q[nw].m+q[nw].s),nw++;
}

for(int i=1;i<=m;i++)
{
if(ans[i]) printf("TAK\n");
else printf("NIE\n");
}

return 0;
}