• PAT T1011 Cut Rectangles


    大模拟题,按要求建立多边形,先定位斜边的位置,再分类讨论~

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn=1014;
    struct node {
        double x,y;
    }Node[2][maxn];
    double distance (node a,node b) {
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }
    int isXiebian (node a,node b) {
        if (a.x!=b.x&&a.y!=b.y) return 1;
        return 0;
    }
    double angle (node a,node b) {
        return fabs(a.x-b.x)/fabs(a.y-b.y);
    }
    vector<double> edge[2];
    int main () {
        int T;
        scanf ("%d",&T);
        int k1,k2;
        while (T--) {
            edge[0].clear();edge[1].clear();
            scanf ("%d",&k1);
            for (int i=0;i<k1;i++) scanf ("%lf %lf",&Node[0][i].x,&Node[0][i].y);
            scanf ("%d",&k2);
            for (int i=0;i<k2;i++) scanf ("%lf %lf",&Node[1][i].x,&Node[1][i].y);
            double xie1=0,xie2=0,cnt1=0,cnt2=0,angle1,angle2,p1,p2;
            for (int i=0;i<k1;i++) {
                int j=i<k1-1?i+1:0;
                if (isXiebian(Node[0][i],Node[0][j])) xie1=distance(Node[0][i],Node[0][j]),cnt1++,
                angle1=angle(Node[0][i],Node[0][j]);
                edge[0].push_back(distance(Node[0][i],Node[0][j]));
            }
            for (int i=0;i<k2;i++) {
                int j=i<k2-1?i+1:0;
                if (isXiebian(Node[1][i],Node[1][j])) xie2=distance(Node[1][i],Node[1][j]),cnt2++,
                angle2=angle(Node[1][i],Node[1][j]);
                edge[1].push_back(distance(Node[1][i],Node[1][j]));
            }
            if (k1+k2>8||k1+k2<6) printf ("NO
    ");
            else if (xie1==xie2&&cnt1==1&&cnt2==1) {
                if ((angle1==angle2||angle1*angle2==1)&&(k1!=4||k2!=4)) printf ("YES
    ");
                else if (k1==4&&k2==4) {
                    double z1=p1<k1-2?p1+2:p1+2-k1;
                    double z2=p2<k2-2?p2+2:p2+2-k2;
                    double z3=p1<k1-1?p1+1:0;
                    double z4=p1>0?p1-1:k1-1;
                    double z5=p2<k2-1?p2+1:0;
                    double z6=p2>0?p2-1:k2-1;
                    if (edge[0][z1]==edge[0][z2]&&(edge[0][z3]+edge[1][z5]==edge[0][z4]+edge[1][z6]||
                    edge[0][z3]+edge[1][z6]==edge[0][z4]+edge[1][z5])) printf ("YES
    ");
                    else printf ("NO
    "); 
                }
                else printf ("NO
    ");
            }
            else if (xie1==0&&xie2==0&&k1==4&&k2==4) {
                for (int i=0;i<k1;i++) {
                    for (int k=0;k<k2;k++) 
                    if (edge[0][i]==edge[1][k]) cnt1++;
                }
                if (cnt1==8||cnt1==16) printf ("YES
    ");
                else printf ("NO
    ");
            } 
            else printf ("NO
    ");
        } 
        return 0;
    } 
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  • 原文地址:https://www.cnblogs.com/zhanglichen/p/12302868.html
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