[练习]: 并查集&最小生成树练习题 2017-06-06 18:47 45人阅读 评论(0) 收藏

嗯。。都是些最最最基础的水题 直接模板就OK了。。 刚学的可以一做

1.洛谷 并查集模板

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 10005
#define M 200005
using namespace std;
int n,m,p;
int fa[M];
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
void solve(){
	scanf("%d%d",&n,&m);
	int k,x,y;
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++){
		scanf("%d%d%d",&k,&x,&y);
		if(k==1)
		merge(x,y);
		else {
			if(findf(x)==findf(y))
			printf("Y
");
			else printf("N
");
		}
	}
}
int main(){
	solve();
	return 0;
} 

2.vijos1034家族

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 5005
#define M 5005 
using namespace std;
int n,m,p;
int fa[M];
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	if(findf(x)!=findf(y)) 
	fa[findf(x)]=findf(y);
}
void solve(){
	scanf("%d%d%d",&n,&m,&p);
	int x,y;
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		merge(x,y);
	}
	for(int i=1;i<=p;i++){
		scanf("%d%d",&x,&y);
		if(findf(x)==findf(y)) printf("Yes
");
		else printf("No
");
	}
}
int main(){
	solve();
	return 0;
} 

3.洛谷p1892团伙

#include<cstdio>
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
int n,m;
int flag;
int f[2500];
int find(int x)
{
    return f[x]==x?x:f[x]=find(f[x]);
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n*2;i++)
    f[i]=i; 
    for(int i=1;i<=m;i++) 
    {
        char t;int x,y;
        cin>>t>>x>>y;
        if(t=='F')      f[find(x)]=find(y);
        if(t=='E')
        {
            f[find(x+n)]=find(y); 
            f[find(y+n)]=find(x);
        }
    }
    int s=0;
    for(int i=1;i<=n;i++)
    if(f[i]==i) s++;
    printf("%d",s);
    return 0;
} 

4.洛谷p2078朋友

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 20005
#define M 20005
using namespace std;
int n,m,p,q;
int a[N],b[N];
int findf1(int x) {
	return a[x]==x?x:a[x]=findf1(a[x]);
}
int findf2(int x) {
	return b[x]==x?x:b[x]=findf2(b[x]);
}
void merge1(int x,int y) {
	if(findf1(x)<findf1(y)) swap(x,y);
	a[findf1(x)]=findf1(y);
}
void merge2(int x,int y) {
	if(findf2(x)<findf2(y)) swap(x,y);
	b[findf2(x)]=findf2(y);
}
int main() {
	scanf("%d%d%d%d",&n,&m,&p,&q);
	int x,y;
	for(int i=1; i<=n; i++) a[i]=i;
	for(int i=1; i<=m; i++) b[i]=i;
	for(int i=1; i<=p; i++) {
		scanf("%d%d",&x,&y);
		merge1(x,y);
	}
	for(int i=1; i<=q; i++) {
		scanf("%d%d",&x,&y);
		merge2(-x,-y);
	}
	int cnta=0,cntb=0;
	for(int i=1; i<=n; i++) {
		if(findf1(1)==findf1(i)) cnta++;
	}
	for(int i=1; i<=m; i++) {
		if(findf2(1)==findf2(i)) cntb++;
	}
	printf("%d",min(cnta,cntb));
	return 0;
}

5.洛谷p1547Out Of Hay

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define N 2005
#define M 10005 
using namespace std;
int n,m;
int fa[N];
struct node{
	int l,r,w;
}e[M];
bool cmp(node x,node y){
	return x.w<y.w;
}
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++)
		scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].w);
	sort(e+1,e+1+m,cmp);
	int ans=0,tot=0;
	for(int i=1;i<=m;i++){
		int x=e[i].l;int y=e[i].r;
		if(findf(x)!=findf(y))
		{
			merge(x,y);
			ans=e[i].w;
			tot++;
		}
		if(tot==n-1) break;
	}
	printf("%d",ans);
	return 0;
}

6.洛谷p1536村村通

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 1005
#define M 10005
using namespace std;
int n,m;
int fa[N];
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
int main(){
	int tot=0;
	while(1){
        bool flag=0;
		tot=0;
		scanf("%d",&n);
		if(n==0) break;
		scanf("%d",&m);
		memset(fa,0,sizeof(fa));
		for(int i=1;i<=n;i++)
			fa[i]=i;
		int x,y;
		for(int i=1;i<=m;i++){
			scanf("%d%d",&x,&y);
			if(findf(x)!=findf(y)&&flag==0)
			{
				merge(x,y);
				tot++;
			}
			if(tot==n-1)
                flag=1;
		}
		printf("%d
",n-1-tot);
	}
	return 0;
}

7.洛谷p1396 营救(最小生成树+并查集)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
#define N 10005
#define M 20005
using namespace std;
int n,m,s,t;
int fa[N];
struct node{
	int l,r,w;
}e[M];
void read(){
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++)
		scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].w);
}
bool cmp(node xx,node yy){
	return xx.w<yy.w;
}
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
int main(){
	int ans=0,cnt=0;
	read();
	sort(e+1,e+1+m,cmp);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++){
		int x=e[i].l;
		int y=e[i].r;
		int w=e[i].w;
		if(findf(x)!=findf(y)) {
			cnt++;
			merge(x,y);
			ans=w;
		}
		if(findf(s)==findf(t)) break;
		if(cnt==n-1) break;
	}
	printf("%d",ans);
	return 0;
}

8.洛谷p1195口袋的天空

/*
P1195 口袋的天空
n个点 m条边 连成k个树的最小代价  
若k为1 则连成一棵MST 跑一次kruskal n-1次连边  
若 k为2 为最后的图必须为 两颗MST n-2次连边
同理 连成k个mst 进行 n-k次连边
若进行者n-k次操作中 没有了边 （即cpp中的 i<m) 直接输出 No Answer
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 1005
#define M 10005
using namespace std;
int n,m,k;
int fa[N];
struct node{
	int l,r,w;
}e[M];
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
void read(){
	scanf("%d%d%d",&n,&m,&k);
	int x,y,t;
	for(int i=1;i<=m;i++)
		scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].w);
}
bool cmp(node xx,node yy){
	return xx.w<yy.w;
}
int main(){
	read();
	for(int i=1;i<=n;i++)
		fa[i]=i;
	sort(e+1,e+1+m,cmp);
	int tot=0,ans=0;
	for(int i=1;i<=m;i++){
		int x=e[i].l;
		int y=e[i].r;
		int w=e[i].w;
		if(findf(x)!=findf(y)){
			merge(x,y);
			tot++;
			ans+=w;
		}
		if(tot==n-k){
			printf("%d",ans);
			return 0;
		}
	}
	printf("No Answer");
	return 0;
}

9.洛谷P1111修复公路

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 1005
#define M 100005
using namespace std;
int n,m;
int fa[N];
struct node{
	int l,r,t;
}e[M];
int findf(int x){
	return fa[x]==x?x:fa[x]=findf(fa[x]);
}
void merge(int x,int y){
	fa[findf(x)]=findf(y);
}
void read(){
	scanf("%d%d",&n,&m);
	int x,y,t;
	for(int i=1;i<=m;i++)
		scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].t);
}
bool cmp(node xx,node yy){
	return xx.t<yy.t;
}
int main(){
	read();
	sort(e+1,e+1+m,cmp);
	int tot=0;
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++){
		int x=e[i].l;
		int y=e[i].r;
		int t=e[i].t;
		if(findf(x)!=findf(y)){
			merge(x,y);
			tot++;
		}
		if(tot==n-1){
			printf("%d",t);
			return 0;
		}
	}
	printf("-1");
	return 0;
}