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

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

瞎搞題，可以用線段樹+lazy過。CB曾經出過一個類似的，可以0(N)的處理。左邊加右邊減，邊走邊算。

#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 3.1415926535898const 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

這道題目感覺挺好的，求在兩個直線之間的所有線段的交點。一開始的想法是左右分別排序根據差值算出交點，但是排序好像有問題。後來改成按右邊的y值排序再求出逆序數就可以了啊。

#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;}

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

這個題目是瞎猜了一下，把它轉化為兩個互質的模型後求出來百分比，用總的長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 3.1415926535898const 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很小可以暴力判斷每次操作之後所在地區的最大值，然後加上h就是總的結果。最後求出一個最大值就可以了。

#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 3.1415926535898const 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