（省赛训练系列）贪心的说 poj贪心经典题目

1
4
1 2 5 10

17

H:过河问题
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1005];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);

        sort(a,a+n);

        if(n<=2)
        {
             printf("%dn",a[n-1]);
             continue;
        }

        int sum=a[1];//最后一定是 a[0].a[1]回去

        while(n>3)
        {
            if(a[0]+a[n-1]+a[0]+a[n-2]>=a[0]+a[1]+a[n-1]+a[1])
                sum+=a[0]+a[1]+a[n-1]+a[1];
            else
                sum+=a[0]+a[n-1]+a[0]+a[n-2];
            n-=2;
        }
        if(n==3)
            sum+=(a[0]+a[2]);
        printf("%dn",sum);
    }
}

Appoint description:  System Crawler  (2015-04-04)

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct ac
{
    double l;
    double r;
}p[1500];

bool cmp(ac x,ac y)
{
    return (x.r!=y.r)?(x.r<y.r):(x.l>y.l);
}

int main()
{
    int n,rad;
    int d=1;
    while(~scanf("%d%d",&n,&rad),(n&&rad))
    {
        int x,y;
        int f=1;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            if(rad>=y)
            {
                p[i].l=x-sqrt((double)rad*rad-(double)y*y);
                p[i].r=x+sqrt((double)rad*rad-(double)y*y);
            }
            else
            {
                f=0;
            }
        }
        if(!f)
        {
            printf("Case %d: -1n",d++);
            continue;
        }
        sort(p,p+n,cmp);

        double End=p[0].r;
        int cnt=1;
        for(int i=1;i<n;i++)
        {
            if(End<p[i].l)
            {
                cnt++;
                End=p[i].r;
            }
        }
        printf("Case %d: %dn",d++,cnt);
    }
}

2 5
1 7 2 10 9

6 11
62 63 54 66 65 61 57 56 50 53 48

0 0

Case 1: 2
Case 2: 4

3 
5 
4 9 5 2 2 1 3 5 1 4 
3 
2 2 1 1 2 2 
3 
1 3 2 2 3 1 

2
1
3

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

struct Rat
{
    int x,y;//l w
} p[5005];
bool vis[5005];

bool cmp(Rat a,Rat b)
{
    return (a.x==b.x)?(a.y<b.y):(a.x<b.x);
}

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d%d",&p[i].x,&p[i].y);

        memset(vis,0,sizeof(vis));

        sort(p,p+n,cmp);

        int cnt=0;
//        int End=p[0].y;
        for(int i=0; i<n; ++i)
        {
            int End=p[i].y;
            if(!vis[i])
            {
                for(int j=i+1; j<n; ++j)
                {
                    if(!vis[j]&&End<=p[j].y)
                    {
                        vis[j]=1;
                        End=p[j].y;
                    }
                }
                cnt++;
            }
        }
        printf("%dn",cnt);
    }
}

4 5
88 200
89 400
97 300
91 500

126900

D：酸奶dp
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int n,s;
int c[10005],y[10005];
int main()
{
    while(~scanf("%d%d",&n,&s))
    {
        for(int i=0;i<n;++i)
            scanf("%d%d",&c[i],&y[i]);
        for(int i=1;i<n;++i)
            c[i]=min(c[i],c[i-1]+s);
        long long ans=0;
        for(int i=0;i<n;i++)
            ans+=(c[i]*y[i]);
        printf("%lldn",ans);
    }

Appoint description:  System Crawler  (2015-04-04)

Description

My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though. 

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size. 

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input

One line with a positive integer: the number of test cases. Then for each test case:

• One line with two integers N and F with 1 ≤ N, F ≤ 10 000: the number of pies and the number of friends.
• One line with N integers ri with 1 ≤ ri ≤ 10 000: the radii of the pies.

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10  ⁻³.

Sample Input

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327
3.1416
50.2655

pie 最大最小值问题 二分
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
double v[10005];
#define esp 1e-6
#define pi acos(-1.0)

int n,f;

double mid_sort(double l,double r)
{
    double mid;
    while(r-l>esp)
    {
        int cnt=0;
        mid=(l+r)/2;
        for(int i=0;i<n;i++)
        {
            cnt+=(int)v[i]/mid;
        }
        if(cnt<f)
            r=mid;
        else
            l=mid;
    }
    return mid;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&f);
        f++;
        double Ms=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf",&v[i]);
            v[i]*=v[i];
            if(Ms<v[i])
                Ms=v[i];
        }
        double k=mid_sort(0.0,Ms);
        printf("%.4lfn",k*pi);
    }
}

25 5 2
2
14
11
21
17

4

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

int a[50005];

int l,n,m;

int b_s(int l,int r)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        int cnt=0,last=0;
        for(int i=1; i<=n+1; i++)
            if(mid>=a[i]-a[last]) cnt++;
            else last=i;
        if(cnt>m)
            r=mid-1;
        else
            l=mid+1;
    }
    return l;
}

int main()
{
    scanf("%d%d%d",&l,&n,&m);
    a[0]=0;
    a[n+1]=l;
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    sort(a,a+n+1);
    printf("%dn",b_s(0,l));

}

679 -> 378 -> 168 -> 48 -> 32 -> 6.

0
1
4
7
18
49
51
768
-1

10
11
14
17
29
77
There is no such number.
2688