pku 1328 第一周訓練 ——貪心

http://poj.org/problem?id=1328

這個題目就是每個島嶼對應一個雷達區間，然后確定好雷達區間后，然后在將區間的s或者e從小到大排序，然后貪心。

第一種按s從小到大排序：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 1007
using namespace std;
struct node
{
    int x,y;
}tagp[maxn];
struct mode
{
    double s,e;
}tagq[maxn];
int cmp(mode a,mode b)
{
    return a.s < b.s;
}

int main()
{
    //freopen("d.in","r",stdin);
    int i,n,d;
    int cas = 1;
    while (scanf("%d%d",&n,&d))
    {
        if (!n && !d) break;

        for (i = 0; i < n; ++i)
        scanf("%d%d",&tagp[i].x,&tagp[i].y);
        int flag = 0;
        for (i = 0; i < n; ++i)
        {
            if (tagp[i].y > d)
            {
               flag = 1;
               break;
            }
            double tmp = sqrt((1.0*d*d) - (1.0*tagp[i].y*tagp[i].y));
            double s = 1.0*tagp[i].x - tmp;
            double e = 1.0*tagp[i].x + tmp;
            tagq[i].s = s;
            tagq[i].e = e;

        }
        if (flag)
        {
             printf("Case %d: %d\n",cas++,-1);
             continue;
        }
        sort(tagq,tagq + n,cmp);
        double e = tagq[0].e;
        int count = 1;
        for (i = 1; i < n; ++i)
        {
            if (tagq[i].s > e)
            {
                count++;
                e = tagq[i].e;
            }
            else
            {
                if (tagq[i].e < e)//注意新的終點的確定
                {
                    e = tagq[i].e;
                }
            }
        }
        printf("Case %d: %d\n",cas++,count);

    }
    return 0;
}

第二中是按e從小到大排序，這里定義int型來接受double型的數據，悲催死我了。。檢查了很長時間。。就是經典貪心之會場安排，不過還是有點差別的。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 1007
using namespace std;
struct node
{
    int x,y;
}tagp[maxn];
struct mode
{
    double s,e;
}tagq[maxn];
bool hash[maxn];
int cmp(mode a,mode b)
{
    return a.e < b.e;
}
int main()
{
    //freopen("d.in","r",stdin);
    int i,count;
    int n,d;
    int cas = 1;
    while (scanf("%d%d",&n,&d))
    {
        if (!n && !d) break;
        int flag = 0;
        for (i = 0; i < n; ++i)
        {
            scanf("%d%d",&tagp[i].x,&tagp[i].y);
            if (tagp[i].y > d)
            {
                flag = 1;
            }
            double tmp = sqrt((1.0*d*d) - 1.0*(tagp[i].y*tagp[i].y));
            tagq[i].s = 1.0*tagp[i].x - tmp;
            tagq[i].e = 1.0*tagp[i].x + tmp;
        }
        if (flag)
        {
            printf("Case %d: %d\n",cas++,-1);
            continue;
        }
        sort(tagq,tagq + n,cmp);
        //for (i = 0; i < n; ++i)
        //printf("%.2lf %.2lf\n",tagq[i].s,tagq[i].e);
        int mark;
        memset(hash,false,sizeof(hash));
        count = 0;
        int j;
        j = 0;
        while (1)
        {
            mark = 0;
            double e = tagq[j].e;
            for(i = 0; i < n; ++i)
            {
                //printf(">>%.2lf %.2lf\n",tagq[i].s,e);
                if (!hash[i] && tagq[i].s <= e)//只有起點小於這個區間的終點的才能公用一個雷達
                {
                    hash[i] = true;
                }
            }
            count++;
            for (i = 0; i < n; ++i)
            {
                if (!hash[i])
                {
                    j = i;
                    mark = 1;
                    break;
                }
            }
            if (!mark)
            {
                printf("Case %d: %d\n",cas++,count);
                break;
            }
        }
    }
    return 0;
}