第一周 2.29 --- 3.6

2.29

cf 620 e New Year Tree

dfs序，线段树区间修改，颜色不超过60种

改了好久，还是写得太少

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<vector>
  6 using namespace std;
  7
  8 typedef long long LL;
  9 const int maxn = 5e6+5;
 10 int a[maxn],b[maxn],ql,qr,qv;
 11 LL sum[maxn],add[maxn];
 12 int vis[maxn],id[maxn],last[maxn],cnt;
 13 vector<int> g[maxn];
 14 int n,m;
 15
 16 void dfs(int u){
 17     id[u] = ++cnt;
 18     for(int i = 0;i < g[u].size();i++){
 19         int v = g[u][i];
 20         if(id[v]) continue;
 21         dfs(v);
 22     }
 23     last[u] = cnt;
 24 }
 25
 26 void Push_up(int o){
 27     sum[o] = sum[o<<1]|sum[o<<1|1];
 28 }
 29
 30 void Push_down(int o){
 31     if(add[o]){
 32         add[o<<1] = add[o<<1|1] = add[o];
 33         sum[o<<1] = 1LL << (add[o]);
 34         sum[o<<1|1] = 1LL << (add[o]);
 35         add[o] = 0LL;
 36     }
 37 }
 38
 39 void Build(int o,int l,int r){
 40     if(l == r){
 41         sum[o] = 1LL << (b[l]);
 42         return;
 43     }
 44     int mid = (l+r)/2;
 45     Build(o<<1,l,mid);
 46     Build(o<<1|1,mid+1,r);
 47     Push_up(o);
 48 }
 49
 50 void Update(int o,int l,int r){
 51     if(ql <= l && r <= qr){
 52         add[o] = qv*1LL;
 53         sum[o] = 1LL << qv;
 54         return;
 55     }
 56     Push_down(o);
 57     int mid = (l+r)/2;
 58     if(ql <= mid) Update(o<<1,l,mid);
 59     if(qr > mid) Update(o<<1|1,mid+1,r);
 60     Push_up(o);
 61 }
 62
 63 LL Query(int o,int l,int r){
 64     if(ql <= l && r <= qr){
 65         return sum[o];
 66     }
 67     Push_down(o);
 68     LL ans = 0LL;
 69     int mid = (l+r)/2;
 70     if(ql <= mid) ans |= Query(o<<1,l,mid);
 71     if(qr > mid) ans |= Query(o<<1|1,mid+1,r);
 72     return ans;
 73 }
 74
 75 void solve(){
 76     memset(add,0,sizeof(add));
 77     Build(1,1,n);
 78     int cmd,c,v;
 79     for(int i = 1;i <= m;i++){
 80         scanf("%d",&cmd);
 81         if(cmd == 1){
 82             scanf("%d %d",&v,&c);
 83             ql = id[v],qr = last[v],qv = c;
 84          //   printf("ql = %d qr = %d qv = %d\n",ql,qr,qv);
 85             Update(1,1,n);
 86         }
 87         else{
 88             scanf("%d",&v);
 89             LL res = 0LL;
 90             ql = id[v],qr = last[v];
 91             res = Query(1,1,n);
 92          //   printf("i = %d res = %I64d\n",i,res);
 93             int zz = 0;
 94             for(int i = 1;i <= 61;i++){
 95                 if(res & (1LL<<i)) zz++;
 96             }
 97             printf("%d\n",zz);
 98         }
 99     }
100 }
101
102 int main(){
103     while(scanf("%d %d",&n,&m) != EOF){
104         for(int i = 1;i <= n;i++) g[i].clear();
105         for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
106         int u,v;
107         for(int i = 1;i <= n-1;i++){
108             scanf("%d %d",&u,&v);
109             g[u].push_back(v);
110             g[v].push_back(u);
111         }
112         memset(vis,0,sizeof(vis));
113         cnt = 0;
114         dfs(1);
115        // for(int i = 1;i <= n;i++) printf("id[%d] = %d last[%d] = %d\n",i,id[i],i,last[i]);
116        memset(b,0,sizeof(b));
117        for(int i = 1;i <= n;i++) b[id[i]] = a[i];
118        solve();
119     }
120     return 0;
121 }

cf 633d 633D - Fibonacci-ish

好神奇

又wa又t的

再用一个map存下已经枚举过的对

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<map>
 6 using namespace std;
 7
 8 typedef long long LL;
 9 int a[1005],n;
10 map<LL,int> h;
11 map<pair<int,int>,int> hh;
12 LL f[1005];
13 int ans,len;
14
15 void dfs(int d){
16     f[d] = f[d-1]+f[d-2];
17     if(h[f[d]] <= 0){
18         len = d-1;
19         return;
20     }
21     h[f[d]]--;
22     dfs(d+1);
23     h[f[d]]++;
24 }
25
26 void solve(){
27     sort(a+1,a+n+1);
28     ans = 2;
29     hh.clear();
30     for(int i = 1;i <= n;i++){
31         for(int j = 1;j <= n;j++){
32             if(i == j) continue;
33             if(hh[make_pair(a[i],a[j])] != 0) continue;
34             hh[make_pair(a[i],a[j])] = 1;
35             f[1] = 1LL*a[i];f[2] = 1LL*a[j];
36
37             h[f[1]]--;h[f[2]]--;
38             len = 2;
39             dfs(3);
40          //   printf("f[1] = %I64d f[2] = %I64d len = %d\n",f[1],f[2],len);
41             ans = max(ans,len);
42             h[f[1]]++;h[f[2]]++;
43         }
44     }
45     printf("%d\n",ans);
46 }
47
48 int main(){
49     while(scanf("%d",&n) != EOF){
50         h.clear();
51         for(int i = 1;i <=n;i++){
52             scanf("%d",&a[i]);
53             h[1LL*a[i]]++;
54         }
55         solve();
56     }
57     return 0;
58 }