BZOJ1483 [HNOI2009]梦幻布丁

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因为是把所有颜色x都变成颜色y，所以把颜色y变成颜色x也是可以的

那么每次呢，就是要把两种颜色合并

记录一下每个颜色当前存在哪个链表里，启发式合并链表就好了

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<vector>#include<map>#include<set>#include<bitset>#include<queue>#include<stack>using namespace std;#define MAXN 100010#define MAXM 1000010#define INF 1000000000#define MOD 1000000007#define eps 1e-8#define ll long longint n,m;int a[MAXN];vector<int>ps[MAXM];int ans;int c[MAXM];int main(){int i;int o,x,y;scanf("%d%d",&n,&m);for(i=1;i<MAXM;i++){c[i]=i;}for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]!=a[i-1]){ans++;}ps[a[i]].push_back(i);}while(m--){scanf("%d",&o);if(o==1){scanf("%d%d",&x,&y);if(x==y){continue ;}if(ps[c[x]].size()>ps[c[y]].size()){swap(c[x],c[y]);}x=c[x];y=c[y];for(i=0;i<ps[x].size();i++){if(a[ps[x][i]-1]==y){ans--;}if(a[ps[x][i]+1]==y){ans--;}}for(i=0;i<ps[x].size();i++){a[ps[x][i]]=y;ps[y].push_back(ps[x][i]);}ps[x].clear();}if(o==2){printf("%d\n",ans);}}return 0;}/**/

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