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  • 数论 矩阵交集

    参考网址 https://www.cnblogs.com/yzm10/p/10963692.html

     输入

    2
    2
    0 0 1 1
    1 1 2 2
    3
    0 0 2 6
    1 1 4 4
    1 3 3 9
    输出
    0 0
    0 0
    2 6
    2 5
    2 5

    求矩阵的交集个数和交集面积
    画出两个矩阵的交集即可,在草稿纸上画出,找出横纵坐标的大小关系,进行比较
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    struct Node{
        int x1,y1,x2,y2;
    }a[5005];
    
    int calc(int a1x1,int a1y1,int a1x2,int a1y2,int a2x1,int a2y1,int a2x2,int a2y2){
        int c=0;
        ll m1=max(min(a1x1,a1x2),min(a2x1,a2x2));
        ll n1=max(min(a1y1,a1y2),min(a2y1,a2y2));
        ll m2=min(max(a1x1,a1x2),max(a2x1,a2x2));
        ll n2=min(max(a1y1,a1y2),max(a2y1,a2y2));
        if(m2>m1&&n2>n1) return 1;
        return 0;
    }
    ll cals(int a1x1,int a1y1,int a1x2,int a1y2,int a2x1,int a2y1,int a2x2,int a2y2){
        int c=0;
        ll m1=max(min(a1x1,a1x2),min(a2x1,a2x2));
        ll n1=max(min(a1y1,a1y2),min(a2y1,a2y2));
        ll m2=min(max(a1x1,a1x2),max(a2x1,a2x2));
        ll n2=min(max(a1y1,a1y2),max(a2y1,a2y2));
        if(m2>m1&&n2>n1) return (m2-m1)*(n2-n1);
        return 0;
    }
    int main()
    {
        int t,n,i,j;
        scanf("%d",&t);
        while(t--){
            scanf("%d",&n);
            for(i=1;i<=n;i++){
                scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
            }
            for(i=1;i<=n;i++){
                int c=0;
                ll s=0;
                for(j=1;j<=n;j++){
                    if(i==j) continue;
                    c+=calc(a[i].x1,a[i].y1,a[i].x2,a[i].y2,a[j].x1,a[j].y1,a[j].x2,a[j].y2);
                    s+=cals(a[i].x1,a[i].y1,a[i].x2,a[i].y2,a[j].x1,a[j].y1,a[j].x2,a[j].y2);
                }
                printf("%d %lld
    ",c,s);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/someonezero/p/12818919.html
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