二分+贪心(总结)

最近做了几道二分+贪心的题目，做下总结

基本思路：通过二分，将范围逐步缩小，直到最优解

1、poj2456

/*题意：    有n个牛栏，选m个放进牛，相当于一条线段上有 n 个点，选取 m 个点，使得相邻点之间的最小距离值最大思路：贪心+二分    二分枚举相邻两牛的间距，判断大于等于此间距下能否放进所有的牛。*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 1e6+10;int a[N],n,m;bool judge(int k)//枚举间距k，看能否使任意两相邻牛{    int cnt = a[0], num = 1;//num为1表示已经第一头牛放在a[0]牛栏中    for(int i = 1; i < n; i ++)//枚举剩下的牛栏    {        if(a[i] - cnt >= k)//a[i]这个牛栏和上一个牛栏间距大于等于k，表示可以再放进牛        {            cnt = a[i];            num ++;//又放进了一头牛        }        if(num >= m) return true;//所有牛都放完了    }    return false;}void solve(){    int l = 1, r = a[n-1] - a[0];//最小距离为1，最大距离为牛栏编号最大的减去编号最小的    while(l < r)    {        int mid = (l+r) >> 1;        if(judge(mid)) l = mid + 1;        else r = mid;    }    printf("%d\n",r-1);}int main(){    int i;    while(~scanf("%d%d",&n,&m))    {        for(i = 0; i < n; i ++)            scanf("%d",&a[i]);        sort(a, a+n);//对牛栏排序        solve();    }    return 0;}

2、poj1064 对浮点数二分

/*题意：    给n条线段，单位为米，要对这些线段裁剪，剪出m条等长的线段，且使这些线段尽可能地长，结果要精确到厘米，即小数点后两位。不能小于1厘米，小于1厘米要输出0.00思路：二分*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 1e6+10;const double eps = 1e-6;double a[N],maxx;int n,m;void solve(){    double l = 0, r = maxx;    double ans = 0;    while(r - l > eps)    {        double mid = (l+r)/2.0;        int sum = 0;        for(int i = 0; i < n; i ++)            sum += (int)(a[i]/mid);//计算能分成多少段        if(sum >= m)            l = mid;        else            r = mid;    }    printf("%.2lf\n",int(r*100)*0.01);//直接输出r的话会四舍五入}int main(){    int i;    while(~scanf("%d%d",&n,&m))    {        maxx = 0;        for(i = 0; i < n; i ++)        {            scanf("%lf",&a[i]);            if(a[i] > maxx) maxx = a[i];        }        solve();    }    return 0;}

3、nyoj914最大化平均值

思路：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 10010;const double eps = 1e-6;int c[N], v[N],n,m;double s[N];bool judge(double x){    int i;    for(i = 0; i < n; i ++)        s[i] = v[i] - x*c[i];    sort(s, s+n);    double sum = 0;    for(i = 0; i < m; i ++)//选s[i]较大的        sum += s[n-1-i];    return sum >= 0;}void solve(){    double l = 0, r = 1000000;    while(r - l > eps)    {        double mid = (l+r)/2;        if(judge(mid))            l = mid;        else            r = mid;    }    printf("%.2lf\n",r);}int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i = 0; i < n; i ++)            scanf("%d%d",&c[i],&v[i]);        solve();    }    return 0;}

4、hdu4004/nyoj619(青蛙过河)

/*题意：    宽为L的河，有n块石头，青蛙可以通过石头跳到河对岸去，最多跳m次，问青蛙每次最少跳多远思路：    假设河的两岸都是石头，一共跳m次，一共有m+1块石头被用到那么我们就可以转化为在n个石头中挑出m+1个石头来解*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 500010;int a[N];int L,n,m;bool judge(int k){    int cnt = 1, pre = a[0];//cnt为1表示已选第一块石头    for(int i = 1; i < n; i ++)    {        if(a[i] - pre > k)        {            pre = a[i-1];            cnt ++;            if(a[i] - pre > k)//两个相邻石头距离大于k                return 0;        }    }    cnt ++;    if(cnt > m + 1)        return 0;    return 1;}void solve(){    int l = 0, r = L;    while(l < r)    {        int mid = (l+r) >> 1;        if(judge(mid))            r = mid;        else            l = mid + 1;    }    printf("%d\n",l);}int main(){    while(~scanf("%d%d%d",&L,&n,&m))    {        a[0] = 0; n ++;        for(int i = 1; i < n; i ++)            scanf("%d",&a[i]);        sort(a+1, a+n);        a[n++] = L;//把河对岸当做最后一个石头        solve();    }    return 0;}

5、nyoj680（摘枇杷）

 #include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 1005;int a[N];int n,m,sum;bool judge(int k){    int cnt = 0, sum = 0;    for(int i = 0; i < n; i ++)    {        if(a[i] > k) return false;        sum += a[i];        if(sum > k)        {            sum = a[i];            cnt ++;        }    }    cnt ++;    if(cnt <= m) return true;    return false;}void solve(){    int l = 0, r = sum;    while(l < r)    {        int mid = (l+r) >> 1;        if(judge(mid))            r = mid;        else            l = mid + 1;    }    printf("%d\n",l);}int main(){    while(~scanf("%d%d",&n,&m))    {        sum = 0;        for(int i = 0; i < n; i ++)            scanf("%d",&a[i]), sum += a[i];        solve();    }    return 0;}