hdu 4368 Water World I

多校联合第七场的题，开始后50分钟第二个ac。简单的计算几何，思路比较明确，求出倾斜后的坐标，扫描一遍所有点，去掉那些水下的点，求一次面积，减去原来面积就行了。复杂度O（n）

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const double Pi=acos(-1.0);const double EP=1e-8;int n, h[10005], vis[20015];double angle;struct Point{    double x, y;}p[20015], pt, tt;double x_mult(Point sp, Point ep, Point op){    return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);}Point intersect(Point u1,Point u2,Point v1,Point v2){      Point res=u1;      double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/               ((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));     res.x+=(u2.x-u1.x)*t;     res.y+=(u2.y-u1.y)*t;     return res;}double area(Point pp[], int sum){    double s=0;    if(sum<3)    return s;    int j;    for(int i=1; i<sum-1;){        while(i<sum-1&&vis[i])i++;        if(i>=sum-1)break;        j=i+1;        while(j<sum&&vis[j])j++;        if(j>=sum)break;        s+=x_mult(pp[i],pp[j],pp[0]);        i=j;    }    s/=2;    return s;}int main(){    //freopen("1.txt", "r", stdin);    int i, j;    while(scanf("%d%lf", &n, &angle)!=EOF){        memset(vis, 0, sizeof(vis));        angle=angle/180.0*Pi;        double s=0, ans=0;        p[0].x=0;p[0].y=0;        for(i=1; i<=n; i++){            scanf("%d", &h[i]);            s+=h[i];            p[2*i-1].x=i-1;p[2*i-1].y=h[i];            p[2*i].x=i;p[2*i].y=h[i];        }        p[2*n+1].x=n;p[2*n+1].y=0;        for(i=1; i<=2*n+1; i++){            pt.x=p[i].x*cos(angle)-p[i].y*sin(angle);            pt.y=p[i].x*sin(angle)+p[i].y*cos(angle);            p[i]=pt;        }        pt=p[0];        int tmp=0;        for(i=1; i<=2*n+1; i++){            if(tmp==i-1&&p[i].y>=pt.y){                tmp=i;                pt=p[i];            }            else if(i<2*n+1&&p[i].y<=pt.y&&p[i+1].y>pt.y+EP){                for(j=tmp+1; j<i; j++)vis[j]=1;                tt.x=pt.x+1;tt.y=pt.y;                p[i]=intersect(pt, tt, p[i], p[i+1]);                pt=p[i];                tmp=i;            }        }        pt=p[2*n+1];tmp=2*n+1;        for(i=2*n; i>=0; i--){            if(tmp==i+1&&p[i].y>=pt.y){                tmp=i;                pt=p[i];            }            else if(i>=1&&p[i].y<=pt.y&&p[i-1].y>pt.y+EP){                for(j=tmp-1; j>i; j--)vis[j]=1;                tt.x=pt.x+1;tt.y=pt.y;                p[i]=intersect(pt, tt, p[i], p[i-1]);                pt=p[i];                tmp=i;            }        }        ans=fabs(area(p, 2*n+2));        printf("%.2f\n", ans-s);    }    return 0;}