题目:
http://www.lydsy.com/JudgeOnline/problem.php?id=2816
思路:
第一个条件看完暂时还没什么想法
看完第二个,发现每一个颜色都是一个森林
进而想到对于每一个颜色维护LCT
看数据范围,n<=10000, c<=10,可行
对于操作0,把每一个LCT上该店的权值都修改
对于操作1,先检测这条边是否存在,若不存在就No such edge
如果这条边存在,且原来的颜色不同于要修改的颜色的话,就先判断Error 1和Error 2
如果要改的边已经是目标颜色了,就特判跳过
上述判断都通过了就cut和link即可
对于操作2,每一个节点维护一个最大值,输出即可
Code:
1 /************************************************************** 2 Problem: 2816 3 User: dedicatus545 4 Language: C++ 5 Result: Accepted 6 Time:8032 ms 7 Memory:6688 kb 8 ****************************************************************/ 9 10 #include<iostream> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cstring> 14 #include<unistd.h> 15 #define col(i,c) n*c+i 16 using namespace std; 17 inline int read(){ 18 int re=0,flag=1;char ch=getchar(); 19 while(ch>‘9‘||ch<‘0‘){ 20 if(ch==‘-‘) flag=-1; 21 ch=getchar(); 22 } 23 while(ch>=‘0‘&&ch<=‘9‘) re=(re<<1)+(re<<3)+ch-‘0‘,ch=getchar(); 24 return re*flag; 25 } 26 struct edge{ 27 int to,next,w; 28 }e[200010]; 29 int n,m,C,cnt=-1,q,first[100010],color[10010][11]; 30 int fa[100010]={0},ch[100010][2]={0},w[100010],a[100010]; 31 bool rev[100010]={0},rt[100010]={0}; 32 void _swap(int &l,int &r){l^=r;r^=l;l^=r;} 33 void update(int x){a[x]=max(a[ch[x][0]],max(a[ch[x][1]],w[x]));}; 34 void pushrev(int x){ 35 if(!x) return; 36 _swap(ch[x][0],ch[x][1]); 37 rev[x]^=1; 38 } 39 void pushdown(int x){ 40 if(!x) return; 41 if(rev[x]){ 42 pushrev(ch[x][0]); 43 pushrev(ch[x][1]); 44 rev[x]=0; 45 } 46 } 47 void push(int x){ 48 //cout<<"push "<<x<<‘ ‘<<rt[x]<<‘\n‘; 49 if(!x) sleep(1000); 50 if(!rt[x]) push(fa[x]); 51 pushdown(x); 52 } 53 int get(int x){return ch[fa[x]][1]==x;} 54 void rotate(int x){ 55 //cout<<"rotate "<<x<<‘\n‘; 56 int f=fa[x],ff=fa[f],son=get(x); 57 ch[f][son]=ch[x][son^1]; 58 if(ch[f][son]) fa[ch[f][son]]=f; 59 ch[x][son^1]=f;fa[f]=x; 60 fa[x]=ff; 61 if(rt[f]) rt[x]=1,rt[f]=0; 62 else ch[ff][ch[ff][1]==f]=x; 63 update(f);update(x); 64 } 65 void splay(int x){ 66 //cout<<"splay "<<x<<‘\n‘; 67 push(x); 68 for(int f;!rt[x];rotate(x)) 69 if(!rt[f=fa[x]]) 70 rotate((get(x)==get(f))?f:x); 71 update(x); 72 } 73 void access(int x){ 74 //cout<<"access "<<x<<‘\n‘; 75 int y=0; 76 do{ 77 splay(x); 78 rt[ch[x][1]]=1; 79 rt[ch[x][1]=y]=0; 80 x=fa[y=x]; 81 update(x); 82 }while(x); 83 } 84 void makeroot(int x){ 85 //cout<<"makeroot "<<x<<‘\n‘; 86 access(x);splay(x);pushrev(x); 87 } 88 bool judge(int x,int y){ 89 while(fa[x]) x=fa[x]; 90 while(fa[y]) y=fa[y]; 91 return x==y; 92 } 93 void init(int x,int y){ 94 //cout<<"init "<<x<<‘ ‘<<y<<‘\n‘; 95 makeroot(x);fa[x]=y; 96 } 97 int link(int x,int y){ 98 if(judge(x,y)) return 0; 99 makeroot(x);fa[x]=y; 100 return 1; 101 } 102 int cut(int x,int y){ 103 if(!judge(x,y)) return 0; 104 makeroot(x);splay(y); 105 fa[ch[y][0]]=fa[y]; 106 rt[ch[y][0]]=1; 107 fa[y]=0;ch[y][0]=0; 108 return 1; 109 } 110 void add(int u,int v,int w){ 111 e[++cnt]=(edge){v,first[u],w};first[u]=cnt; 112 e[++cnt]=(edge){u,first[v],w};first[v]=cnt; 113 } 114 int split(int u,int v){ 115 if(!judge(u,v)) return -1; 116 makeroot(u);access(v);splay(v); 117 return a[v]; 118 } 119 int main(){ 120 // freopen("networkzj.in","r",stdin); 121 // freopen("networkzj.out","w",stdout); 122 memset(first,-1,sizeof(first)); 123 int i,j,t1,t2,t3,t4; 124 n=read();m=read();C=read();q=read(); 125 for(i=1;i<=n;i++){ 126 t1=read(); 127 for(j=0;j<C;j++) a[col(i,j)]=w[col(i,j)]=t1,rt[col(i,j)]=1; 128 } 129 130 for(i=1;i<=m;i++){ 131 t1=read();t2=read();t3=read(); 132 init(col(t1,t3),col(t2,t3)); 133 add(t1,t2,t3); 134 color[t1][t3]++;color[t2][t3]++; 135 } 136 //for(i=1;i<=n;i++){ 137 //for(j=0;j<C;j++) cout<<color[i][j]<<‘ ‘; 138 //cout<<‘\n‘; 139 //} 140 for(i=1;i<=q;i++){ 141 //cout<<"operation "<<i<<‘\n‘; 142 t1=read(); 143 if(t1==0){ 144 t2=read();t3=read(); 145 for(j=0;j<C;j++){ 146 makeroot(col(t2,j));w[col(t2,j)]=t3;update(col(t2,j)); 147 } 148 } 149 if(t1==1){ 150 t2=read();t3=read();t4=read();bool flag=1,f=1; 151 for(j=first[t2];~j;j=e[j].next) 152 if(e[j].to==t3){ 153 flag=0; 154 if(e[j].w==t4) f=0; 155 } 156 if(flag){ 157 printf("No such edge.\n"); 158 continue; 159 } 160 if(f&&(color[t2][t4]==2||color[t3][t4]==2)){ 161 printf("Error 1.\n");continue; 162 } 163 if(f&&(judge(col(t2,t4),col(t3,t4)))){ 164 printf("Error 2.\n");continue; 165 } 166 for(j=first[t2];~j;j=e[j].next){ 167 if(e[j].to==t3){ 168 color[t2][e[j].w]--;color[t3][e[j].w]--; 169 cut(col(t2,e[j].w),col(t3,e[j].w)); 170 e[j].w=e[j^1].w=t4; 171 color[t2][t4]++;color[t3][t4]++; 172 link(col(t2,t4),col(t3,t4)); 173 printf("Success.\n");break; 174 } 175 } 176 } 177 if(t1==2){ 178 t2=read();t3=read();t4=read(); 179 printf("%d\n",split(col(t3,t2),col(t4,t2))); 180 } 181 } 182 }
原文地址:https://www.cnblogs.com/dedicatus545/p/8387511.html
时间: 2024-11-13 08:17:44