【最小生成树】【并查集】[USACO2016 金组]Fenced In

题目描述

Farmer John has realized that many of his cows are strangely agoraphobic (being fearful of large open spaces). To try and make them less afraid of grazing, he partitions his large field into a number of smaller regions by building vertical (north-south) and horizontal (east-west) fences.
The large field is a rectangle with corner points at (0,0) and (A,B). FJ builds n vertical fences (0≤n≤2000) at distinct locations a1…ana1…an (0<ai

INPUT FORMAT (file fencedin.in):

The first line of input contains AA, BB, nn, and mm (1≤A,B≤1,000,000,000). The next nn lines contain a1…an, and the next mm lines after that contain b1…bm.

OUTPUT FORMAT (file fencedin.out):

Please write the minimum length of fencing FJ must remove. Note that this might be too large to fit into a standard 32-bit integer, so you may need to use 64-bit integer types (e.g., “long long” in C/C++).

SAMPLE INPUT:

15 15 5 2
2
5
10
6
4
11
3

SAMPLE OUTPUT:

44

题目分析

很显然给每个方格编号相邻的连接边的权值为边的长度，但是因为边的数量太大，我们不能直接存储，我们发现每一行的所有边的长度都相同，同理每一列的长度也相同，所以我们有只用存储每一行的长度加边的时候在遍历该长度所在的行或者列进行连边，然后同时我们用启发式合并的方法进行合并并查集能够胜任本题目。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const int MAXN = 2001 * 2001;
int Fa[MAXN+10], n, m, Rank[MAXN+10];
int A, B;
int As[2010], Bs[2010];
typedef pair<int, pair<int, int> > plii;
typedef pair<int, int> pii;
plii lis[2010*2+10];
char ch;
void Read(int &s){
    while((ch=getchar()) < '0' || ch > '9');
    s = ch - '0';
    while((ch=getchar()), ch>='0'&&ch<='9')
        s = s*10+ch-'0';
    ungetc(ch, stdin);
}
int findFa(int u){
    if(!Fa[u]) return u;
    return Fa[u]=findFa(Fa[u]);
}
inline int GetID(int x, int y){return (n+1)*x + y;}
int main(){
    //printf("%.3lf\n", 1.0*(sizeof(lis) +sizeof(Fa))/1024/1024);
    Read(A); Read(B); Read(n); Read(m);
    for(int i=1;i<=n;++i) Read(As[i]);
    sort(As+1, As+1+n);
    for(int i=1;i<=m;++i) Read(Bs[i]);
    sort(Bs+1, Bs+1+m);
    Bs[m+1] = B; As[n+1] = A;
    int cnt = 0;
    for(int i=0;i<=m;++i) lis[cnt++] = plii(Bs[i+1]-Bs[i],pii(-1, i));
    for(int i=0;i<=n;++i) lis[cnt++] = plii(As[i+1]-As[i],pii(i, -1));
    long long ans = 0, did = 0, Max = (n+1)*(m+1)-1;
    sort(lis, lis+cnt);
    for(int i=0;i<cnt && did<Max;++i){
        if(lis[i].second.first == -1){
            for(int j=0;j<n;++j){
                int p1=GetID(lis[i].second.second, j);
                int p2=GetID(lis[i].second.second, j+1);
                int f1=findFa(p1), f2=findFa(p2);
                if(f1 == f2) continue;
                if(Rank[f1] > Rank[f2]) swap(f1, f2);
                Fa[f1] = f2;
                Rank[f2] = max(Rank[f1]+1, Rank[f2]);
                ans += 1LL * lis[i].first;
                ++did;
            }
        }else{
            for(int j=0;j<m;++j){
                int p1=GetID(j, lis[i].second.first);
                int p2=GetID(j+1, lis[i].second.first);
                int f1=findFa(p1), f2=findFa(p2);
                if(f1 == f2) continue;
                if(Rank[f1] > Rank[f2]) swap(f1, f2);
                Fa[f1] = f2;
                Rank[f2] = max(Rank[f1]+1, Rank[f2]);
                ans += 1LL * lis[i].first;
                ++did;
            }
        }
    }
    printf("%lld\n", ans);

    return 0;
}