Zoj monthly, February 2012 C, D, F, H

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

You can use the line segment tree + lazy to solve the problem. CB once had a similar process, which can be 0 (n. Subtract the value on the left and calculate the value while walking.

#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 3.1415926535898const 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;}

D: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3574.

This question is quite good. Find the intersection of all line segments between two straight lines. The idea at the beginning was to sort the left and right separately and calculate the intersection point based on the difference value, but the sorting seemed to be problematic. Later, I changed it to sorting by the Y value on the right and then finding the reverse order number.

#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;}

F: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 4598

This question is a guess. convert it into two mutually dependent models and calculate the percentage. Multiply the total length by the percentage.

#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 3.1415926535898const 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;}

H: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3578.

N is very small, it can be used to determine the maximum value of the region after each operation, and then H is the total result. Finally, you can find a maximum value.

#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 3.1415926535898const 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