4863. 【GDOI2017模拟11.5】Market

Description

Data Constraint

Solution

对于60分的解法，我们直接打个背包即可。然后我们发现：M[I]实在太大了，所以我们设出f[i]表示在价值为i的情况下的最小代价，转移方式与原来相同，然后每个询问二分一下答案即可。由于v[i]*n最多只有90000，所以复杂度为O(300n2)。

Code

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=100005,mx=90000;
struct code{
int a,b,c;
}a[maxn],b[maxn];
ll n,m,i,t,j,k,l,r,mid,f[mx+5],ans[maxn],g[mx+5];
bool cmp(code x,code y){
return x.c<y.c;
}
int main(){
freopen("market.in","r",stdin);freopen("market.out","w",stdout);
//freopen("data.in","r",stdin);freopen("data.out","w",stdout);
scanf("%lld%lld",&n,&m);
for (i=1;i<=n;i++)
scanf("%lld%lld%lld",&a[i].a,&a[i].b,&a[i].c);
sort(a+1,a+n+1,cmp);
for (i=1;i<=m;i++)
scanf("%lld%lld",&b[i].c,&b[i].a),b[i].b=i;
sort(b+1,b+m+1,cmp);j=1;
memset(f,127,sizeof(f));f[0]=0;memset(g,127,sizeof(g));
for (i=1;i<=m;i++){
while (a[j].c<=b[i].c && j<=n){
for (k=mx-a[j].b;k>=0;k--)
if (f[k]>=0)f[k+a[j].b]=min(f[k+a[j].b],f[k]+a[j].a);
for (k=mx;k>=0;k--)
g[k]=min(g[k+1],f[k]);
j++;
}
l=0;
r=mx;
while (l<r){
mid=(l+r+1)/2;
if (g[mid]>b[i].a) r=mid-1;
else l=mid;
}
ans[b[i].b]=l;
}
for (i=1;i<=m;i++)
printf("%lld\n",ans[i]);
}