BZOJ 2794 DP

思路：

考虑把询问离线

按照m排序

物品按照a排序

f[i]表示c[j]的和到i b的最大值

背包就好

O(nk)竟然能过……

//By SiriusRen
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=1005;
struct Ask{int m,k,s,id;}ask[N*N];
struct Node{int a,b,c;}node[N];
bool cmp(Ask a,Ask b){return a.m<b.m;}
bool Cmp(Node a,Node b){return a.a<b.a;}
int n,q,a[N],b[N],c[N],jy=1,ans[N*N],f[N*101];
void update(int x){for(int i=N*100-node[x].c;i>=0;i--)f[i+node[x].c]=max(f[i+node[x].c],min(f[i],node[x].b));}
int main(){
    scanf("%d",&n),f[0]=0x3f3f3f3f;
    for(int i=1;i<=n;i++)scanf("%d%d%d",&node[i].c,&node[i].a,&node[i].b);
    scanf("%d",&q);
    for(int i=1;i<=q;i++)scanf("%d%d%d",&ask[i].m,&ask[i].k,&ask[i].s),ask[i].id=i;
    sort(ask+1,ask+1+q,cmp),sort(node+1,node+1+n,Cmp);
    for(int i=1;i<=q;i++){
        while(jy<=n&&node[jy].a<=ask[i].m)update(jy),jy++;
        ans[ask[i].id]=(f[ask[i].k]>(ask[i].m+ask[i].s)?1:0);
    }
    for(int i=1;i<=q;i++)puts(ans[i]?"TAK":"NIE");
}