zoukankan      html  css  js  c++  java
  • BZOJ4424: Cf19E Fairy

    树上差分的代码很简洁,dfs+差分即可

    这题很多坑点啊,比如重边自环好坑

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<vector>
     6 #define pii pair<int,int>
     7 #define pb push_back
     8 #define mp make_pair
     9 #define ft first
    10 #define sc second
    11 #define MAXN 1000000+10
    12 using namespace std;
    13 int first[MAXN],edges[MAXN<<1],nxt[MAXN<<1],to[MAXN<<1];
    14 int n,m,c=-1;
    15 int b[MAXN],d[MAXN],tag[MAXN];
    16 int tmp,cnt,self_tmp;
    17 void insert(int x,int y,int e){
    18     nxt[++c]=first[x],first[x]=c,to[c]=y,edges[c]=e;
    19     nxt[++c]=first[y],first[y]=c,to[c]=x,edges[c]=e;
    20 }
    21 void init(){
    22     scanf("%d%d",&n,&m);
    23     int x,y;
    24     for(int i=1;i<=m;i++){
    25         scanf("%d%d",&x,&y);
    26         if(x==y&&!self_tmp){self_tmp=i;continue;}
    27         if(x==y){self_tmp=-1;continue;}
    28         insert(x,y,i);
    29     }
    30 }
    31 void dfs(int x,int fa){
    32     for(int e=first[x];e;e=nxt[e]){
    33         if(e==(fa^1))continue;
    34         int &y=to[e];
    35         if(d[y]){
    36             if(d[x]<d[y])continue;
    37             if((d[x]-d[y]+1)&1){
    38                 tag[x]++,tag[y]--;
    39                 cnt++;tmp=edges[e];
    40             }
    41             else{
    42                 tag[x]--,tag[y]++;
    43             }
    44         }
    45         else{
    46             d[y]=d[x]+1;
    47             dfs(y,e);
    48             tag[x]+=tag[y];
    49         }
    50     }
    51 }
    52 vector<int> ans;
    53 void find(int x){
    54     b[x]=1;
    55     for(int e=first[x];e;e=nxt[e]){
    56         int &y=to[e];
    57         if(b[y])continue;
    58         if(tag[y]==cnt){
    59             ans.pb(edges[e]);
    60         }
    61         find(y);
    62     }
    63 }
    64 void solve(){
    65     if(-1==self_tmp){printf("0
    ");return;}
    66     for(int i=1;i<=n;i++){
    67         if(!d[i])d[i]=1,dfs(i,0);
    68     }
    69     if(!cnt){
    70         if(self_tmp){printf("1
    %d
    ",self_tmp);}
    71         else{printf("%d
    ",m);for(int i=1;i<m;i++)printf("%d ",i);printf("%d
    ",m);}
    72         return;
    73     }
    74     if(self_tmp){printf("0
    ");return;}
    75     for(int i=1;i<=n;i++){
    76         if(!b[i])find(i);    
    77     }
    78     if(cnt==1)ans.pb(tmp);
    79     sort(ans.begin(),ans.end());
    80     printf("%d
    ",ans.size());
    81     for(int i=0;i<ans.size();i++){
    82         printf("%d",ans[i]);
    83         if(i==ans.size()-1)printf("
    ");
    84         else printf(" ");
    85     }
    86 }
    87 int main()
    88 {
    89 //  freopen("data.in","r",stdin);  
    90 //  freopen("my.out","w",stdout);  
    91     init();
    92     solve();
    93     return 0;
    94 }
    View Code
  • 相关阅读:
    CSDN的博客是肿么了?
    SQL Server 2005: 存储过程签名
    SQLSERVER中数据行所占用的最小空间
    如何配置 SQL Server 2005 以允许远程连接
    XamlPad小程序
    OPEN SYMMETRIC KEY scope in SQL Server
    SQL Server 2005 helpful catalogs: crypt_properties and key_encryptions
    WPF Unleashed Chapter 3:Important New Concepts in WPF Routed Events
    xp_cmdshell
    SQL Server 2005: 如何让用户只能加密数据却不能解密数据
  • 原文地址:https://www.cnblogs.com/w-h-h/p/8246477.html
Copyright © 2011-2022 走看看