來源 codeforces 2020 GDUT Rating Contest II (Div. 2) CF鏈接

題目：
Farmer John’s N cows, conveniently numbered 1…N (2≤N≤1e5), have a complex social structure revolving around “moo networks” — smaller groups of cows that communicate within their group but not with other groups. Each cow is situated at a distinct (x,y) location on the 2D map of the farm, and we know that M pairs of cows (1≤M<1e5) moo at each-other. Two cows that moo at each-other belong to the same moo network.

In an effort to update his farm, Farmer John wants to build a rectangular fence, with its edges parallel to the x and y axes. Farmer John wants to make sure that at least one moo network is completely enclosed by the fence (cows on the boundary of the rectangle count as being enclosed). Please help Farmer John determine the smallest possible perimeter of a fence that satisfies this requirement. It is possible for this fence to have zero width or zero height.

Input
The first line of input contains N and M. The next N lines each contain the x and y coordinates of a cow (nonnegative integers of size at most 108). The next M lines each contain two integers a and b describing a moo connection between cows a and b. Every cow has at least one moo connection, and no connection is repeated in the input.

Output
Please print the smallest perimeter of a fence satisfying Farmer John’s requirements.

Example
input
7 5
0 5
10 5
5 0
5 10
6 7
8 6
8 4
1 2
2 3
3 4
5 6
7 6
output
10

題意：n頭牛在不同的位置，某兩頭牛是一伙的，并且和一伙的其他牛的一伙的也是一伙的，求一最短的柵欄能把其中某一伙牛給框起來。
思路：用并查集來維護，每加入一頭牛到這個集團就更新這個集團的四個角的坐標到集團首，然后枚舉每個集團需要多長的柵欄即可。

代碼:

#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #define INF 0x3f3f3f3f #define mod 1000000007 using namespace std; struct cow {int a,b,c,d; }k[100010]; int n,m,pre[100010]; int find(int x) {int k=x;while (pre[x]!=x)x=pre[x];pre[k]=x;return x; } void join(int a,int b) {int x=find(a),y=find(b);if (x!=y){pre[x]=y;k[y].a=min(k[x].a,k[y].a);k[y].b=max(k[x].b,k[y].b);k[y].c=min(k[x].c,k[y].c);k[y].d=max(k[x].d,k[y].d);} } int main() {cin>>n>>m;for (int i=1;i<=n;i++){scanf("%d%d",&k[i].a,&k[i].c);k[i].b=k[i].a,k[i].d=k[i].c;}int a,b;for (int i=1;i<=n;i++) pre[i]=i;for (int i=1;i<=m;i++){scanf("%d%d",&a,&b);join(a,b);}long long ans=400000001;for (int i=1;i<=n;i++){long long L;int x=find(i);L=2*(k[x].b-k[x].a+k[x].d-k[x].c);ans=min(ans,L);}cout<<ans;return 0; }

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