版权声明:喜欢请点个大拇指,感谢各位dalao。弱弱说下,转载要出处呦 https://blog.csdn.net/qq_35786326/article/details/88872715
题目:
题意:
每次可选择一个点,对连接该点的所有边进行变色
求将整个图变为同一颜色最少需要多少步
分析:
我们可以将此图转换为
染色图,对于红蓝两色,先假设整图为一种,然后对所有边进行变色,再变为另一种颜色
而判断是否可以变色,可以用
来暴力模拟
代码:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<list>
#include<ctime>
#include<iomanip>
#include<string>
#include<bitset>
#include<deque>
#include<set>
#define LL long long
#define king 99999999
using namespace std;
inline LL read(){
LL d=0,f=1;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
return d*f;
}
int n,m,tot,mins,sum,flag;
int ver[200005],next[200005],edge[200005],head[200005],v[100005],f[100005],run[100005];
void add(int x,int y,char w)
{
ver[++tot]=y;
next[tot]=head[x];
head[x]=tot;
edge[tot]=(w=='R');
return;
}
void dfs(int x)
{
v[x]=run[x]=1;
for(int i=head[x];i;i=next[i])
{
if(flag)break;
int y=ver[i];
if(!v[y])
{
f[y]=f[x]^edge[i];
if(f[y])sum++;
dfs(y);
}
else if(f[y]!=(f[x]^edge[i])) flag=1;
}
return;
}
void work()
{
memset(run,0,sizeof(run));
int ans=0,k;
for(int i=1;i<=n;i++)
{
if(run[i]) continue;
f[i]=sum=flag=0;
memset(v,0,sizeof(v));
dfs(i);
if(flag)k=king;
else k=sum;
f[i]=sum=1;
flag=0;
memset(v,0,sizeof(v));
dfs(i);
if(flag)sum=king;
if(k==king&&sum==king)
{
ans=king;
break;
}
ans+=min(k,sum);
}
mins=min(ans,mins);
}
int main()
{
n=read();m=read();
for(int i=1;i<=m;i++)
{
int x,y;
char c;
x=read();y=read();scanf("%c",&c);
add(x,y,c);
add(y,x,c);
}
mins=king;
work();
for(int i=1;i<=tot;i++) edge[i]^=1;
work();
if(mins==king) printf("-1");
else cout<<mins;
return 0;
}