mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
112 lines
2.8 KiB
C++
112 lines
2.8 KiB
C++
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#include <stdio.h>
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#include <string.h>
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#include <iostream>
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#include <algorithm>
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#include <queue>
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#include <map>
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#include <set>
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#include <vector>
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#include <string>
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#include <math.h>
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using namespace std;
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const int MAXN = 200010;
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int x[MAXN],y[MAXN];
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int d;
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int n;
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int L;
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int nowx ;
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int nextx;
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int r1,l2;
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const double eps = 1e-8;
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double solve()
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{
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double left = nowx,right = nextx;
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double ret1,ret2;
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for(int cc = 0;cc <= 30;cc++)
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{
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double mid = (left + right)/2;
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double midmid = (mid + right)/2;
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double h1 = y[r1] + (double)(y[r1-1] - y[r1]) * (mid - x[r1])/(x[r1-1] - x[r1]);
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double h2 = y[l2] + (double)(y[l2+1] - y[l2])*(mid + 2*d - x[l2])/(x[l2 + 1] - x[l2]);
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ret1 = (double)(x[r1] - mid)*(h1 + y[r1])/2 + (double)(mid + 2*d - x[l2])*(h2 + y[l2])/2;
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h1 = y[r1] + (double)(y[r1-1] - y[r1]) * (midmid - x[r1])/(x[r1-1] - x[r1]);
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h2 = y[l2] + (double)(y[l2+1] - y[l2])*(midmid + 2*d - x[l2])/(x[l2 + 1] - x[l2]);
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ret2 = (double)(x[r1] - midmid)*(h1 + y[r1])/2 + (double)(midmid + 2*d - x[l2])*(h2 + y[l2])/2;
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if(ret1 < ret2)
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left = mid+eps;
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else right = midmid-eps;
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}
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return ret1;
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}
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int input()
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{
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char ch;
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ch = getchar();
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while(ch < '0' || ch >'9')
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{
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ch = getchar();
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}
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int ret = 0;
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while(ch >= '0' && ch <= '9')
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{
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ret *= 10;
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ret += ch -'0';
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ch = getchar();
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}
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return ret;
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}
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int main()
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{
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int T;
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scanf("%d",&T);
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while(T--)
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{
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scanf("%d%d",&n,&L);
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for(int i = 1;i <= n;i++)
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{
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scanf("%d%d",&x[i],&y[i]);
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}
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scanf("%d",&d);
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double ans = 0;
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r1 = 2;
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l2 = 1;
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double tmp = 0;
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while(l2 < n && x[l2+1] < 2*d)l2++;
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for(int i = r1;i < l2;i++)
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{
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tmp += (double)(x[i+1] - x[i])*(y[i] + y[i+1])/2;
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}
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if(l2 == 1)
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{
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tmp -= (double)(x[2] - x[1])*(y[2] + y[1])/2;
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}
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x[n+1] = x[n];
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y[n+1] = y[n];
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nowx = 0;
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while(l2 < n && r1 <= n)
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{
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int p1 = x[r1];
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int p2 = x[l2 + 1] - 2*d;
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if(p1 < p2)
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nextx = p1;
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else nextx = p2;
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nextx = min(L- 2*d,nextx);
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ans = max(ans,tmp + solve());
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if(p1 < p2)
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{
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nowx = p1;
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if(r1 < n)tmp -= (double)(x[r1+1] - x[r1])*(y[r1+1] + y[r1] )/2;
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r1++;
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}
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else
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{
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nowx = p2;
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tmp += (double)(x[l2+1] - x[l2])*(y[l2+1] + y[l2])/2;
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l2++;
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}
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}
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printf("%.3lf\n",ans/2/d);
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}
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return 0;
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}
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