论蒟蒻的自我修养T_T。。
和noi2014魔法森林基本一样。。。然而数据范围大得sxbk。。。UPD:这题如果用lct判联通的话可能会被卡到O(mlogm)。。所以最好还是用并查集吧
一开始数组开太小re了两发（要开到maxn+maxm），然后又开太大mle一发，然后无限tle。。。把记录类型全改成数组还是tle。。。。
最后把非lct部分改得和黄学长全部一样终于20+s卡过去了。。。。。。。。。
然后发现自己原来是有个地方写萎了。。一开始把没被删的边做kruskal的时候，只要加入n-1条边那么整张图就联通了。。。所以加了n-1条边后就可以直接break。。
蒟蒻不仅没跳出来，判断是否联通还用lct。。。结果O(mlogm)光荣tle。。。。
加了break就18s过了。。。。。。。。。。。。。。
1 #include<cstdio> 2 #include<iostream> 3 #include<math.h> 4 #include<algorithm> 5 using namespace std; 6 const int maxn=100033; 7 const int maxm=1000033; 8 int treec[maxm+maxn][2],treefa[maxm+maxn],treemax[maxm+maxn],treepos[maxm+maxn],treeval[maxm+maxn]; 9 bool treerev[maxm+maxn]; 10 struct zs{ 11 int id,k,x,y,val; 12 }ask[maxn]; 13 struct edge{ 14 int id,from,too,val; 15 bool broken; 16 }e[maxm],e1[maxm]; 17 int stack[maxm+maxn],ans[maxn]; 18 int i,j,n,m,q; 19 20 inline int read() 21 { 22 int x=0,f=1;char ch=getchar(); 23 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 24 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 25 return x*f; 26 } 27 inline bool isroot(int x){return treec[treefa[x]][0]!=x&&treec[treefa[x]][1]!=x; 28 } 29 inline void update(int x){ 30 int l=treec[x][0],r=treec[x][1]; 31 treemax[x]=treeval[x];treepos[x]=x; 32 if(treemax[l]>treemax[x])treemax[x]=treemax[l],treepos[x]=treepos[l]; 33 if(treemax[r]>treemax[x])treemax[x]=treemax[r],treepos[x]=treepos[r]; 34 } 35 inline void pushdown(int x){ 36 if(!treerev[x])return; 37 int l=treec[x][0],r=treec[x][1]; 38 if(l)treerev[l]^=1;if(r)treerev[r]^=1; 39 treerev[x]^=1;swap(treec[x][0],treec[x][1]); 40 } 41 inline void rotate(int x){ 42 int fa=treefa[x],gfa=treefa[fa]; 43 if(!isroot(fa))treec[gfa][treec[gfa][1]==fa]=x; 44 int l=treec[fa][1]==x,r=l^1; 45 treec[fa][l]=treec[x][r];treec[x][r]=fa; 46 treefa[x]=gfa;treefa[fa]=x;treefa[treec[fa][l]]=fa; 47 update(fa);update(x); 48 } 49 inline void splay(int x){ 50 int top=0,tmp=x;stack[++top]=x; 51 while(!isroot(tmp))stack[++top]=treefa[tmp],tmp=treefa[tmp]; 52 while(top)pushdown(stack[top]),top--; 53 int fa,gfa; 54 while(!isroot(x)){ 55 fa=treefa[x];gfa=treefa[fa]; 56 if(!isroot(fa)) 57 if((treec[gfa][0]==fa)^(treec[fa][0]==x))rotate(x); 58 else rotate(fa); 59 rotate(x); 60 } 61 } 62 inline void access(int x){ 63 int son=0; 64 while(x){ 65 splay(x);treec[x][1]=son;update(x); 66 son=x;x=treefa[x]; 67 } 68 } 69 inline void makeroot(int x){ 70 access(x);splay(x);treerev[x]^=1; 71 } 72 inline void link(int x,int y){ 73 makeroot(x);treefa[x]=y; 74 } 75 inline void cut(int x,int y){ 76 makeroot(x);access(y);splay(y);treec[y][0]=treefa[x]=0;update(y); 77 } 78 inline int getfa(int x){ 79 access(x);splay(x); 80 while(treec[x][0])x=treec[x][0],pushdown(x); 81 return x; 82 } 83 inline int half(int x,int y){ 84 int l=0,r=m,mid; 85 while(l<r){ 86 mid=l+r>>1; 87 if((e[mid].from<x)||(e[mid].from==x&&e[mid].too<y))l=mid+1; 88 else r=mid; 89 } 90 return l; 91 } 92 bool cmp(edge a,edge b){ 93 return ((a.from<b.from)||(a.from==b.from&&a.too<b.too)); 94 } 95 bool cmp1(edge a,edge b){ 96 return a.val<b.val; 97 } 98 inline int getpos(int x,int y){ 99 makeroot(x);access(y);splay(y);return treepos[y]; 100 } 101 int main(){ 102 n=read();m=read();q=read(); 103 for(i=1;i<=m;i++){ 104 e[i].from=read();e[i].too=read();e[i].val=read(); 105 if(e[i].from>e[i].too)swap(e[i].from,e[i].too); 106 } 107 sort(e+1,e+1+m,cmp); 108 for(i=1;i<=m;i++)e[i].id=i,treeval[i+n]=treemax[i+n]=e[i].val,treepos[i+n]=i+n; 109 for(i=1;i<=q;i++){ 110 ask[i].k=read();ask[i].x=read();ask[i].y=read(); 111 if(ask[i].k==2)if(ask[i].x>ask[i].y)swap(ask[i].x,ask[i].y); 112 if(ask[i].k==2)ask[i].id=half(ask[i].x,ask[i].y),e[ask[i].id].broken=1,ask[i].val=e[ask[i].id].val; 113 } 114 sort(e+1,e+1+m,cmp1); 115 for(i=1;i<=m;i++)e1[e[i].id]=e[i]; 116 int tot=0; 117 for(i=1;i<=m;i++) 118 if(!e[i].broken){ 119 int x=e[i].from,y=e[i].too; 120 if(getfa(x)!=getfa(y))link(x,e[i].id+n),link(e[i].id+n,y),tot++; 121 if(tot==n-1)break; 122 } 123 for(i=q;i;i--){ 124 if(ask[i].k==1)ans[i]=treeval[getpos(ask[i].x,ask[i].y)]; 125 else if(ask[i].k==2){ 126 int x=ask[i].x,y=ask[i].y,pos=getpos(x,y); 127 if(treeval[pos]>ask[i].val){ 128 cut(e1[pos-n].from,pos);cut(pos,e1[pos-n].too); 129 link(x,ask[i].id+n);link(ask[i].id+n,y); 130 } 131 } 132 } 133 for(i=1;i<=q;i++)if(ask[i].k==1)printf("%d\n",ans[i]); 134 return 0; 135 }