ZOJ Monthly, February 2012 C,D,F,H

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C:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3573

瞎搞题，可以用线段树+lazy过。CB曾经出过一个类似的，可以0(N)的处理。左边加右边减，边走边算。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898

const int maxn = 15100;

using namespace std;

LL f[maxn];
LL p[maxn];
LL k[maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int x, y, z;
        memset(f, 0, sizeof(f));
        memset(p, 0, sizeof(p));
        memset(k, 0, sizeof(k));
        while(~scanf("%d %d %d",&x, &y, &z))
        {
            if(x == -1) break;
            f[x] += z;
            p[y+1] -= z;
        }
        int lmax;
        LL Max = -1LL;
        LL xmax = 0LL;
        for(int i = 0; i <= n; i++)
        {
            xmax += f[i]+p[i];
            k[i] = xmax;
            if(xmax > Max)
            {
                Max = xmax;
                lmax = i;
            }
        }
        int rmax;
        for(int i = n; i >= 0; i--)
        {
            if(k[i] == Max)
            {
                rmax = i;
                break;
            }
        }
        cout<<lmax<<" "<<rmax<<endl;
    }
    return 0;
}

这道题目感觉挺好的，求在两个直线之间的所有线段的交点。一开始的想法是左右分别排序根据差值算出交点，但是排序好像有问题。后来改成按右边的y值排序再求出逆序数就可以了啊。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 30100;

LL sum;

struct node
{
    LL y1;
    LL y2;
    int id;
};

node a[maxn], temp[maxn];
void Merge(node a[], int l, int mid, int r)
{
    int begin1 = l;
    int end1 = mid;
    int begin2 = mid+1;
    int end2 = r;
    int k = 0;
    while(begin1 <= end1 && begin2 <= end2)
    {
        if(a[begin1].y1 < a[begin2].y1)
        {
            temp[k++] = a[begin1];
            begin1++;
            sum += begin2-(mid+1);
        }
        else
        {
            temp[k++] = a[begin2];
            begin2++;
        }
    }
    while(begin1 <= end1)
    {
        temp[k++] = a[begin1];
        begin1++;
        sum += end2-mid;
    }
    while(begin2 <= end2)
    {
        temp[k++] = a[begin2];
        begin2++;
    }
    for(int i = l; i <= r; i++) a[i] = temp[i-l];
}

void MergeSort(node a[], int l, int r)
{
    int mid = (l+r)>>1;
    if(l < r)
    {
        MergeSort(a, l, mid);
        MergeSort(a, mid+1, r);
        Merge(a, l, mid, r);
    }
}

bool cmp(node a, node b)
{
    return a.y2 < b.y2;
}

int main()
{
    LL x1, x2;
    while(~scanf("%lld %lld",&x1, &x2))
    {
        int n;
        scanf("%d",&n);
        if(n == 0)
        {
            cout<<1<<endl;
            continue;
        }
        LL k, b;
        for(int i = 0; i < n; i++)
        {
            scanf("%lld %lld",&k, &b);
            a[i].y1 = k*x1+b;
            a[i].y2 = k*x2+b;
        }
        sort(a, a+n, cmp);
        sum = 0;
        MergeSort(a, 0, n-1);
        cout<<sum+n+1<<endl;
    }
    return 0;
}

这个题目是瞎猜了一下，把它转化为两个互质的模型后求出来百分比，用总的长度乘以百分比就可以了啊。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898

const int maxn = 30010;

using namespace std;

int main()
{
    LL n, m;
    while(~scanf("%lld %lld",&n, &m))
    {
        double x = sqrt(n*n*1.0+m*m*1.0);
        LL p = __gcd(n, m);
        m /= p;
        n /= p;
        p = m*n;
        LL xx = p/2;
        if(p%2) xx++;
        double y = (xx*1.0)/p;
        printf("%.6lf\n",x*y);
    }
    return 0;
}

n很小可以暴力判断每次操作之后所在区域的最大值，然后加上h就是总的结果。最后求出一个最大值就可以了。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898

const int maxn = 15100;

using namespace std;

struct node
{
    int h;
    int H;
    int a, b, x, y;
}f[maxn];

bool judge(int x, int y)
{
    if(f[x].x+f[x].a <= f[y].x || f[y].x+f[y].a <= f[x].x) return false;
    if(f[x].y+f[x].b <= f[y].y || f[y].y+f[y].b <= f[x].y) return false;
    return true;
}

int main()
{
    int n, m, c;
    while(~scanf("%d %d %d",&n, &m, &c))
    {
        for(int i = 0; i < c; i++)
        {
            scanf("%d %d %d %d %d",&f[i].a, &f[i].b, &f[i].h, &f[i].x, &f[i].y);
            f[i].H = 0;
            for(int k = 0; k < i; k++)
                if(judge(i, k)) f[i].H = max(f[i].H, f[k].H);
            f[i].H += f[i].h;
        }
        int Max = -1;
        for(int i = 0; i < c; i++) Max = max(Max, f[i].H);
        cout<<Max<<endl;
    }
    return 0;
}

ZOJ Monthly, February 2012 C,D,F,H

标签：style   blog   http   io   os   ar   for   sp   on   

原文地址：http://blog.csdn.net/xu12110501127/article/details/40679389