搜索-洛谷P1034 矩形覆盖

来源：互联网发布：大数据架构详解编辑：程序博客网时间：2024/06/01 21:31

https://daniu.luogu.org/problem/show?pid=1034
瞎几把想了半天，想出了一个可以过的暴搜方案，打着打着就萎了；
然后看题解，妈的全是瞎搞，然后4^50的暴搜加上优化有人过了；
然后我就去百度看看；
https://wenku.baidu.com/view/5f8961b47fd5360cba1adbeb.html
然后就知道这个是noip提高的题目；
说好k<=4，其实k最大只有3；
然后数据没有对每一种情况近判断性，所以本来300+的代码因为有些用不到可以变成100+；
我曹；

#include<iostream>#include<cstdio>#include<cstdlib>#define Ll long longusing namespace std;int x[51],y[51];int n,m,x1,x2,y1,y2;void A(){    x1=y1=1e9;x2=y2=0;    for(int i=1;i<=n;i++){        x1=min(x1,x[i]);        x2=max(x2,x[i]);        y1=min(y1,y[i]);        y2=max(y2,y[i]);    }    if(x1==1e9)x1=x2=0;    printf("%d",(x2-x1)*(y2-y1));}void B(){    int ans,sum=1e9;    for(int k=0;k<=500;k++){        ans=0;        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(x[i]>k)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(x[i]<=k)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);        sum=min(sum,ans);    }    for(int k=0;k<=500;k++){        ans=0;        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(y[i]>k)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(y[i]<=k)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);        sum=min(sum,ans);    }    printf("%d",sum);}void C(){    int ans,sum=1e9;    for(int k=0;k<=500;k++)    for(int j=k+1;j<=500;j++){        ans=0;        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(y[i]>k)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);    //  if(k==1&&j==2)cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<endl;        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(k>=y[i]||y[i]>j)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;        ans+=(x2-x1)*(y2-y1);    //  if(k==1&&j==2)cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<endl;        x1=y1=1e9;x2=y2=0;        for(int i=1;i<=n;i++){            if(y[i]<=j)continue;            x1=min(x1,x[i]);            x2=max(x2,x[i]);            y1=min(y1,y[i]);            y2=max(y2,y[i]);        }        if(x1==1e9)x1=x2=0;    //  if(k==1&&j==2)cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<endl;        ans+=(x2-x1)*(y2-y1);        sum=min(sum,ans);    }    printf("%d",sum);}int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);     if(m==1)A();    if(m==2)B();    if(m==3)C();}

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