Codeforces 731C Socks 并查集
2016-11-10 20:48
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C. Socks
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared
a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of
Arseniy's n socks is assigned a unique integer from 1 to n.
Thus, the only thing his mother had to do was to write down two integers li and ri for
each of the days — the indices of socks to wear on the day i (obviously, li stands
for the left foot and ri for
the right). Each sock is painted in one of kcolors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars
with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear
the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during
each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) —
the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2,
..., cn (1 ≤ ci ≤ k) —
current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) —
indices of socks which Arseniy should wear during thei-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
input
output
input
output
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
并查集,合并袜子,最后分别找每一个集合的总数,最大颜色数量,相减的求总和就是答案
可是,数组过不了会超时,用map才过,最后遍历数组谁的颜色集合不空,即证明他是boos,
对于每个boos遍历一遍他的颜色数组,找到最大值
#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<time.h>
using namespace std;
int c[200009];
int f[200009];
int num[200009];
vector <int >v[200009];
int find(int x){
return x==f[x]?x:(f[x]=find(f[x]));
}
void merge(int x,int y){
int a=find(x);
int b=find(y);
if(a!=b){
f[b]=a;
num[a]+=num[b]; //a为集合老大,b集合的数量要加在a上
}
}
int maxx(int a,int b){
return a>b?a:b;
}
int main(){
clock_t t1,t2,t3;
int n,m,k,i,j,l,r,ans,max;
scanf("%d%d%d",&n,&m,&k);
//并查集模板
for(i=1;i<=n;i++){
scanf("%d",c+i);
f[i]=i;
num[i]=1;
}
for(i=1;i<=m;i++){
scanf("%d%d",&l,&r);
merge(l,r);
}
//vector存入,并用find更新一次f
//如boss i的小弟包括i自身的每一个颜色
ans=0;
for(i=1;i<=n;i++)
v[find(i)].push_back(c[i]) ;
//查找每个boss
//map用来存boss的每种颜色的个数,边存边找最大值max
//
//t1=clock()/CLOCKS_PER_SEC ;
for(i=1;i<=n;i++){
if(v[i].size()<=1)continue ;
max=0;
map<int,int> mm;
//int mm[200009];
//memset(mm,0,sizeof(mm));
for(j=0;j<v[i].size() ;j++){
mm[v[i][j]]++;
if(max<mm[v[i][j]]) max=mm[v[i][j]];
}
ans+=(num[i]-max);
}
//t2=clock()/CLOCKS_PER_SEC-t1;
//printf("time=%lf\n",t2);
printf("%d\n",ans);
return 0;
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared
a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of
Arseniy's n socks is assigned a unique integer from 1 to n.
Thus, the only thing his mother had to do was to write down two integers li and ri for
each of the days — the indices of socks to wear on the day i (obviously, li stands
for the left foot and ri for
the right). Each sock is painted in one of kcolors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars
with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear
the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during
each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) —
the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2,
..., cn (1 ≤ ci ≤ k) —
current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) —
indices of socks which Arseniy should wear during thei-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
input
3 2 3 1 2 3 1 2 2 3
output
2
input
3 2 21 1 21 22 1
output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
并查集,合并袜子,最后分别找每一个集合的总数,最大颜色数量,相减的求总和就是答案
可是,数组过不了会超时,用map才过,最后遍历数组谁的颜色集合不空,即证明他是boos,
对于每个boos遍历一遍他的颜色数组,找到最大值
#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<time.h>
using namespace std;
int c[200009];
int f[200009];
int num[200009];
vector <int >v[200009];
int find(int x){
return x==f[x]?x:(f[x]=find(f[x]));
}
void merge(int x,int y){
int a=find(x);
int b=find(y);
if(a!=b){
f[b]=a;
num[a]+=num[b]; //a为集合老大,b集合的数量要加在a上
}
}
int maxx(int a,int b){
return a>b?a:b;
}
int main(){
clock_t t1,t2,t3;
int n,m,k,i,j,l,r,ans,max;
scanf("%d%d%d",&n,&m,&k);
//并查集模板
for(i=1;i<=n;i++){
scanf("%d",c+i);
f[i]=i;
num[i]=1;
}
for(i=1;i<=m;i++){
scanf("%d%d",&l,&r);
merge(l,r);
}
//vector存入,并用find更新一次f
//如boss i的小弟包括i自身的每一个颜色
ans=0;
for(i=1;i<=n;i++)
v[find(i)].push_back(c[i]) ;
//查找每个boss
//map用来存boss的每种颜色的个数,边存边找最大值max
//
//t1=clock()/CLOCKS_PER_SEC ;
for(i=1;i<=n;i++){
if(v[i].size()<=1)continue ;
max=0;
map<int,int> mm;
//int mm[200009];
//memset(mm,0,sizeof(mm));
for(j=0;j<v[i].size() ;j++){
mm[v[i][j]]++;
if(max<mm[v[i][j]]) max=mm[v[i][j]];
}
ans+=(num[i]-max);
}
//t2=clock()/CLOCKS_PER_SEC-t1;
//printf("time=%lf\n",t2);
printf("%d\n",ans);
return 0;
}
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