WUST OJ 1421 we love girl(贪心或DP）

Description

When taking part in ACM Programming Contest,many school hope girls for reception like ccnu,cug and so on.So this year our wust send more girls as possible to have a reception.The ratio of male to female of wust is seven to one,so we may not have so many girls.Of
course the rest receptionist is boys.

Now there are n schools take part in Wust ACM Programming Contest,x girls and y boys is willing to help.If a school is received buy girls,they will get p satisfaction.If received buy boys ,they will get q satisfaction.We want to improve the total satisfaction
of every school.Do you know the max satisfaction?

Input

The first line contains an integer T(T<=100), indicates the number of cases.For each test case,the first line contains three positive integers n (1<=n<=20),x and y (x>=0,y>=0,x+y>=n).n is the amount of schools and x is the amount of girls and y is the amount
of boys.Then n lines follows, each line contains two integers pi and qi which means the two satisfaction of girls and boys.

Output

For each test case,print one integer which is the max satisfaction.

Sample Input

2

3 2 1

2 1

4 2

1 2

4 2 2

3 2

2 4

10 9

4 0

Sample Output

8

20

这个题当时贪心写的。n比较小，发挥不了贪心优势。

由于只有两种选择，选择一种就意味着放弃另一种。

我们贪心的策略就是按照选男孩和女孩差值排序，然后每次

选最优的（如果可能）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=(1e5+100)*2;
struct node{
    int x,y;
    int val;
}e[25];
int t,n,x,y;
int cmp(node l1,node l2)
{
    return l1.val>l2.val;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&x,&y);
        int sum=0,ans=0;
        REPF(i,1,n)
        {
            scanf("%d%d",&e[i].x,&e[i].y);
            e[i].val=abs(e[i].x-e[i].y);
        }
        sort(e+1,e+1+n,cmp);
        for(int i=1;i<=n;i++)
        {
            if(e[i].x>e[i].y&&x)
            {
                ans+=e[i].x;
                x--;
            }
            else if(e[i].x<e[i].y&&y)
            {
                ans+=e[i].y;
                y--;
            }
            else if(e[i].x==e[i].y)
                ans+=e[i].x;
            else if(!x)
                ans+=e[i].y;
            else ans+=e[i].x;
        }
        printf("%d\n",ans);
    }
    return 0;
}

显然也可以DP做。设前i个人选了j个女孩的最大值。

注意如果x>n要令x=n..

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
int t,n,x,y;
int c1[110],c2[110];
int dp[110][110];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&x,&y);
        REPF(i,1,n)
          scanf("%d%d",&c1[i],&c2[i]);
        if(x>n) x=n;
        for(int i=0;i<=n;i++)
           for(int j=0;j<=x;j++) dp[i][j]=-INF;
        for(int i=1;i<=x;i++)
            dp[i][i]=dp[i-1][i-1]+c1[i];
        dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=x;j++)
            {
                if(i-j>y) continue;
                dp[i][j]=dp[i-1][j]+c2[i];
                if(j>0) dp[i][j]=max(dp[i][j],dp[i-1][j-1]+c1[i]);
            }
        }
        int ans=-INF;
        for(int i=0;i<=x;i++)
            ans=max(ans,dp
[i]);
        printf("%d\n",ans);
    }
    return  0;
}

/*

*/