（技巧）AtCoder Grand Contest 018 C : Coins

Problem Statement

There are X+Y+Z people, conveniently numbered 1 through X+Y+Z. Person i has Ai gold coins, Bi silver coins and Ci bronze coins.

Snuke is thinking of getting gold coins from X of those people, silver coins from Y of the people and bronze coins from Z of the people. It is not possible to get two or more different colors of coins from a single person. On the other hand, a person will give all of his/her coins of the color specified by Snuke.

Snuke would like to maximize the total number of coins of all colors he gets. Find the maximum possible number of coins.

Constraints

  • 1≤X
  • 1≤Y
  • 1≤Z
  • X+Y+Z≤105
  • 1≤Ai≤109
  • 1≤Bi≤109
  • 1≤Ci≤109

Input

Input is given from Standard Input in the following format:

X Y Z
A1 B1 C1
A2 B2 C2
:
AX+Y+Z BX+Y+Z CX+Y+Z

Output

Print the maximum possible total number of coins of all colors he gets.

Sample Input 1

Copy
1 2 1
2 4 4
3 2 1
7 6 7
5 2 3

Sample Output 1

Copy
18

Get silver coins from Person 1, silver coins from Person 2, bronze coins from Person 3 and gold coins from Person 4. In this case, the total number of coins will be 4+2+7+5=18. It is not possible to get 19 or more coins, and the answer is therefore 18.

Sample Input 2

Copy
3 3 2
16 17 1
2 7 5
2 16 12
17 7 7
13 2 10
12 18 3
16 15 19
5 6 2

Sample Output 2

Copy
110

Sample Input 3

Copy
6 2 4
33189 87907 277349742
71616 46764 575306520
8801 53151 327161251
58589 4337 796697686
66854 17565 289910583
50598 35195 478112689
13919 88414 103962455
7953 69657 699253752
44255 98144 468443709
2332 42580 752437097
39752 19060 845062869
60126 74101 382963164

Sample Output 3

Copy
3093929975

用（ai-ci,bi-ci，0）替换（ai,bi,ci）的想法非常精彩。排序时的考虑也值得思考。
 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <vector>
12 #include <stack>
13 #define mp make_pair
14 #define MIN(a,b) (a>b?b:a)
15 //#define MAX(a,b) (a>b?a:b)
16 typedef long long ll;
17 typedef unsigned long long ull;
18 const int MAX=1e5+5;
19 const int INF=1e8+5;
20 using namespace std;
21 //const int MOD=1e9+7;
22 typedef pair<ll,int> pii;
23 const double eps=0.00000001;
24 struct node
25 {
26     ll a,b;
27 }re[MAX];
28 bool cmp(node x,node y)
29 {
30     if(x.a-x.b!=y.a-y.b)
31         return x.a-x.b<y.a-y.b;
32     else
33         return x.b>y.b;
34 }
35 ll x,y,z;
36 ll gold[MAX],silver[MAX];
37 ll sum,gsum,ssum;
38 priority_queue<ll> g,s;
39 int main()
40 {
41     scanf("%lld%lld%lld",&x,&y,&z);
42     ll num=x+y+z;
43     ll p,q,r;
44     for(ll i=1;i<=num;i++)
45     {
46         scanf("%lld%lld%lld",&p,&q,&r);
47         re[i].a=p-r;re[i].b=q-r;
48         sum+=r;
49     }
50     sort(re+1,re+1+num,cmp);
51     ssum=gsum=0LL;
52     for(ll i=x+y+z;i>=y+z+1;i--)
53     {
54         g.push(-re[i].a);
55         gsum+=re[i].a;
56     }
57     gold[x]=gsum;
58     for(ll i=y+z;i>y;i--)
59     {
60         g.push(-re[i].a);
61         gsum+=re[i].a;
62         gsum+=g.top();
63         gold[x+y+z-i+1]=gsum;
64         g.pop();
65     }
66     for(ll i=1;i<=y;i++)
67     {
68         s.push(-re[i].b);
69         ssum+=re[i].b;
70     }
71     silver[y]=ssum;
72     for(ll i=y+1;i<=y+z;i++)
73     {
74         s.push(-re[i].b);
75         ssum+=re[i].b;
76         ssum+=s.top();
77         silver[i]=ssum;
78         s.pop();
79     }
80     ll an;
81     for(ll i=y;i<=y+z;i++)
82     {
83         if(i!=y)
84             an=max(an,silver[i]+gold[x+y+z-i]);
85         else
86             an=silver[i]+gold[x+y+z-i];
87     }
88     printf("%lld\n",an+sum);
89     return 0;
90 }

 









            

            
posted @ 
2017-07-24 00:15 
perplex 
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