|洛谷|并查集|P1525 关押罪犯

最新推荐文章于 2025-03-30 13:55:02 发布

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最新推荐文章于 2025-03-30 13:55:02 发布

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分类专栏：洛谷树 - 并查集

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本文介绍了罪犯分配问题的三种解决方法：拆点法、边分离法及带权并查集。通过C++代码实现，详细展示了每种方法的具体操作步骤。

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https://www.luogu.org/problem/show?pid=1525

方法一：拆点，即一个点拆为一个实结点和一个虚结点

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define ms(i,j) memset(i,j, sizeof i);
using namespace std; 
struct node
{
	int x;
	int y;
	int v;
	bool operator <(node const &a) const//重载符号以排序 
	{
		return v>a.v;
	}
}zf[100005];//罪犯 
int n,m;
int father[40005];//并查集 
int find(int x)//并查集找根+路径压缩 
{
	if (father[x]==x) return x;
	return father[x] = find(father[x]);
}
int merge(int x, int y)//并查集合并 
{
	int x1 = find(x);
	int y1 = find(y);
	father[y1] = x1;
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i=1;i<=2*n;i++) father[i] = i; 
	for (int i=1;i<=m;i++)	scanf("%d%d%d", &zf[i].x, &zf[i].y, &zf[i].v);//输入 
	sort(zf+1, zf+1+m);//排序 
	for (int i=1;i<=m;i++)
	{
		int x1 = find(zf[i].x);
		int y1 = find(zf[i].y);
		if (x1!=y1)//不在同一监狱 
		{
			merge(zf[i].x, zf[i].y+n);
			merge(zf[i].y, zf[i].x+n);//合并虚结点 
		} else {printf("%d\n", zf[i].v);return 0;}
	}
	printf("%d\n", 0);
    return 0;
}

方法二：先按权值从大到小排序边然后依次使边的两个点分开

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
struct ed
{
    int x,y,v;
}edge[100005];
int father[20005];
int find(int x)
{
    if (father[x]!=x) return father[x] = find(father[x]);
    else return father[x];
}
int merge(int x, int y)
{
    father[x] = y;
}
bool cmp(ed const &a, ed const &b)
{
    return a.v>b.v;
}
int n,m;
int last[20005];//上一个敌人
int main()
{
    scanf("%d%d", &n ,&m);
    for (int i=1;i<=n;i++) father[i] = i;
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].v);
    }
    sort(edge+1, edge+1+m, cmp);
    ms(last,-1);
    for (int i=1;i<=m;i++)
    {
        int a = find(edge[i].x);
        int b = find(edge[i].y);
        if (a==b) 
        {
            printf("%d\n", edge[i].v);
            return 0;
        }
        if (last[b]>0) merge(a,find(last[b]));
        if (last[a]>0) merge(b,find(last[a]));
        last[edge[i].x] = b, last[edge[i].y] = a;
    }
    printf("0\n");
    return 0;
}

方法三：带权并查集

#include<cstdio>    
#include<algorithm>    
#include<cstring>    
#include<queue>    
#define ms(i,j) memset(i,j, sizeof i);    
using namespace std;     
struct node    
{    
    int x;    
    int y;    
    int v;    
    bool operator <(node const &a) const//重载符号以排序     
    {    
        return v>a.v;    
    }    
}zf[100005];//罪犯     
int n,m;    
int father[20005];//并查集  
int r[20005];//指出的权值   1表示在同一监狱 0表示不在同一监狱 
int find(int x)//并查集找根+路径压缩     
{    
    if (father[x]==x) return x;  
	int tmp = father[x];  
    father[x] = find(father[x]);
    r[x] = (r[x]+r[tmp])%2; 
    return father[x];    
}    
int main()    
{    
    scanf("%d%d", &n, &m);    
    for (int i=1;i<=n;i++) father[i] = i;     
    for (int i=1;i<=m;i++)   scanf("%d%d%d", &zf[i].x, &zf[i].y, &zf[i].v);//输入     
    sort(zf+1, zf+1+m);//排序     
    for (int i=1;i<=m;i++)    
    {    
       int a = zf[i].x, b = zf[i].y, x = find(a), y = find(b);
       if (x!=y)
       {
			father[x] = y;
			r[x] = (r[b]-r[a]+1+666)%2;
	   } else
	   {
	   		if ((r[b]-r[a]+1)%2) {printf("%d\n", zf[i].v);   return 0;} 
	   }
    }    
    printf("%d\n", 0);    
    return 0;    
}