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CF 3B Lorry

2015-07-09 11:18 274 查看
sort + greedy

为了满足各种判断条件 这样做应该是比较优的

全都用1的 全都用2的 1 2混合使用的

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAX = 1e5 + 1000;

struct bb
{
int t,p,num;
} kay[MAX],aca[MAX];
int cmp(const bb & a,const bb & b)
{
return a.p > b.p;
}
int n,v,num[MAX],sum_num;
int main()
{
while(~scanf("%d%d",&n,&v))
{
int t,p;
int num_kay = 0,num_aca = 0;
for(int i = 0; i < n; i ++)
{
scanf("%d%d",&t,&p);
if(t%2 == 0)
{
aca[num_aca].t = t;
aca[num_aca].p = p;
aca[num_aca].num = i + 1;
num_aca ++;
}
else
{
kay[num_kay].t = t;
kay[num_kay].p = p;
kay[num_kay].num = i + 1;
num_kay ++;
}
}
sort(kay,kay+num_kay,cmp);
sort(aca,aca+num_aca,cmp);
LL sum1[MAX],sum2[MAX];
sum1[0] = 0;
sum2[0] = 0;
for(int i = 1; i <= num_kay; i ++)
sum1[i] = sum1[i-1] + (LL)kay[i-1].p;
for(int i = 1; i <= num_aca; i ++)
sum2[i] = sum2[i-1] + (LL)aca[i-1].p;

int h1 = 0,h2 = 0,p1,p2;
LL ans = 0,sum = 0;
for(int i = 0; i <= num_kay; i ++)
{
if(i > v) break;
int rest = v - i;
if(rest >= num_aca*2) p2 = num_aca;
else p2 = rest /2;
sum = sum1[i] + sum2[p2];
if(sum > ans)
{
ans = sum;
h1 = i;
h2 = p2;
}
}
cout<<ans<<endl;
sum_num = 0;
for(int i = 0;i < h2;i ++)
{
num[sum_num ++] = aca[i].num;
}
for(int i = 0;i < h1;i ++)
{
num[sum_num ++] = kay[i].num;
}
sort(num,num+sum_num);
for(int i = 0;i < sum_num;i ++)
cout<<num[i]<<" ";
cout<<endl;

}
return 0;
}
/**
3 2
1 2
2 7
1 3
sum1  3 2
sum2  7
4 3
1 4
2 6
1 1
1 1.
sum1 4 1 1
sum2 6
2 2
1 4
2 6
sum1 4
sum2 6
*/
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