Title three hetero Atcoder
T1 Integers on a Tree
Effect: to give you a tree, a start has been determined at some point of the weight is required in O (n) time, it is determined whether there is an arrangement such that the weight of the weight of the program and other adjacent points differ by exactly one, output program
Solution: It can be observed from the extended limit outward like a point known to limit the reach-point must be the tolerance of the arithmetic series 2, so again traversing the entire tree, the processing of each point out restriction i.e. may need maintenance operations merging two arithmetic sequence (taking an intersection), and the arithmetic sequence that alterations
1 #include<bits/stdc++.h> 2 #define FOR(i,a,b) for (register int i=(a);i<=(b);i++) 3 #define For(i,a,b) for (register int i=(a);i>=(b);i--) 4 #define mem(i,j) memset(i,j,sizeof(i)) 5 #define GO(u) for (register int j=f[u];j!=-1;j=nxxt[j]) 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define pii pair<int,int> 10 #define MP make_pair 11 using namespace std; 12 typedef long long ll; 13 const int N=2e5+5; 14 int n,tmp1,tmp2,k,vv[N],p[N],col[N],tag[N],ans[N]; 15 int tot=0,f[N],nxxt[N<<1]; 16 struct E 17 { 18 int u,v; 19 }e[N<<1]; 20 inline void add(int u,int v) 21 { 22 tot++; 23 nxxt[tot]=f[u]; 24 f[u]=tot; 25 e[tot]=(E){u,v}; 26 return; 27 } 28 inline int read() 29 { 30 int x=0,f=1; 31 char c=getchar(); 32 while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} 33 while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();} 34 return f*x; 35 } 36 inline void write(int x) 37 { 38 if (x<0) putchar('-'),x=-x; 39 if (x>9) write(x/10); 40 putchar(x%10+'0'); 41 return; 42 } 43 struct data 44 { 45 int first,siz; 46 inline data work() 47 { 48 data ret; 49 ret.first=first-1; 50 ret.siz=siz+1; 51 return ret; 52 } 53 inline void merge(data x) 54 { 55 if (first==-1&&siz==-1) {first=x.first;siz=x.siz;return;} 56 if (((x.first+2*n)%2)!=((first+2*n)%2)) {first=0;siz=0;return;} 57 int min1=first,max1=first+(siz-1)*2,min2=x.first,max2=x.first+(x.siz-1)*2; 58 int mn=max(min1,min2),mx=min(max1,max2); 59 if (mn>mx) {first=0,siz=0;return;} 60 first=mn; 61 siz=(mx-mn)/2+1; 62 return; 63 } 64 inline void Print() 65 { 66 write(first),putchar(' '),write(siz),putchar('\n'); 67 return; 68 } 69 }s[N]; 70 inline void no() 71 { 72 printf("No\n"); 73 exit(0); 74 } 75 inline data dfs(int u,int fa) 76 { 77 if (s[u].siz==0) no(); 78 data tmp,ret; 79 data nxt=s[u].work(); 80 GO(u) 81 { 82 int v=e[j].v; 83 if (v==fa) continue; 84 s[v].merge(nxt); 85 tmp=dfs(v,u); 86 s[u].merge(tmp); 87 if (s[u].siz==0) no(); 88 } 89 ret=s[u].work(); 90 return ret; 91 } 92 inline void debug() 93 { 94 FOR(i,1,n) 95 { 96 printf("%d : ",i); 97 s[i].Print(); 98 } 99 return; 100 } 101 inline void confirm(int u,int fa) 102 { 103 if (s[u].first==-1) s[u]=(data){1,1}; 104 ans[u]=s[u].first; 105 s[u].siz=1; 106 data nxt=s[u].work(); 107 GO(u) 108 { 109 int v=e[j].v; 110 if (v==fa) continue; 111 s[v].merge(nxt); 112 confirm(v,u); 113 } 114 return; 115 } 116 int main() 117 { 118 mem(f,-1); 119 n=read(); 120 FOR(i,1,n-1) tmp1=read(),tmp2=read(),add(tmp1,tmp2),add(tmp2,tmp1); 121 k=read(); 122 FOR(i,1,k) vv[i]=read(),p[i]=read(),col[vv[i]]=p[i],tag[vv[i]]=1,s[vv[i]]=(data){p[i],1}; 123 FOR(i,1,n) if (!tag[i]) s[i]=(data){-1,-1}; 124 dfs(vv[1],0); 125 confirm(vv[1],0); 126 printf("Yes\n"); 127 FOR(i,1,n) write(ans[i]),putchar('\n'); 128 return 0; 129 }
T2 Shik and Travel
Effect: to a binary tree, all non-leaf son number two, the search from the root of the tree again, minimizing path and a leaf from a leaf to the next maximum
answer:
- Minimize the maximum, half the answer
- In all the dot recording state of each point, etc., from the bottom up merging, each state is a tuple (a, B), represents the distance between the first point down to a leaf of a, a last leaf Back this distance is the point b, and the other two state values to be combined
- Due to excessive number of states, must be considered inferior to abandon the existing state of Excellence status for (a, b) and (c, d), if we put a and c pieces together, we get (a, d), empathy another obtained when (c, b), the former as an example, for a fixed a, in (b + c) two symbols separated like weights, we wish to minimize is d, then combined when the first sort, then two sweep the pointers on the line
1 #include<bits/stdc++.h> 2 #define FOR(i,a,b) for (register ll i=(a);i<=(b);i++) 3 #define For(i,a,b) for (register ll i=(a);i>=(b);i--) 4 #define mem(i,j) memset(i,j,sizeof(i)) 5 #define GO(u) for (register ll j=f[u];j!=-1;j=nxt[j]) 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define pii pair<ll,ll> 10 #define MP make_pair 11 using namespace std; 12 typedef long long ll; 13 const ll N=150000; 14 ll n,a[N],v[N],mn[20000005],L,R,MID,du[N]; 15 vector <pii> F[N]; 16 ll tot=0,f[N],nxt[N<<1]; 17 struct E 18 { 19 ll u,v,w; 20 }e[N<<1]; 21 inline void add(ll u,ll v,ll w) 22 { 23 tot++; 24 nxt[tot]=f[u]; 25 f[u]=tot; 26 e[tot]=(E){u,v,w}; 27 return; 28 } 29 inline ll read() 30 { 31 ll x=0,f=1; 32 char c=getchar(); 33 while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} 34 while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();} 35 return f*x; 36 } 37 inline void write(ll x) 38 { 39 if (x<0) putchar('-'),x=-x; 40 if (x>9) write(x/10); 41 putchar(x%10+'0'); 42 return; 43 } 44 bool cmpfi(const pii x,const pii y) {return x.fi<y.fi;} 45 bool cmpse(const pii x,const pii y) {return x.se<y.se;} 46 inline void update(vector<pii> &G,ll ls,ll rs) 47 { 48 ll siz1=F[ls].size(),siz2=F[rs].size(); 49 if ((!siz1)||(!siz2)) return; 50 if (siz1>siz2) swap(ls,rs),swap(siz1,siz2); 51 sort(F[ls].begin(),F[ls].end(),cmpfi); 52 sort(F[rs].begin(),F[rs].end(),cmpse); 53 mn[0]=F[rs][0].fi; 54 FOR(i,1,siz2-1) mn[i]=min(mn[i-1],F[rs][i].fi); 55 for (register ll i=0,pt=siz2-1;i<siz1;i++) 56 { 57 while (pt>-1&&F[ls][i].fi+F[rs][pt].se>MID) pt--; 58 if (pt>=0) G.pb(MP(mn[pt],F[ls][i].se)); 59 } 60 sort(F[ls].begin(),F[ls].end(),cmpse); 61 sort(F[rs].begin(),F[rs].end(),cmpfi); 62 mn[0]=F[rs][0].se; 63 FOR(i,1,siz2-1) mn[i]=min(mn[i-1],F[rs][i].se); 64 for (register ll i=0,pt=siz2-1;i<siz1;i++) 65 { 66 while (pt>-1&&F[ls][i].se+F[rs][pt].fi>MID) pt--; 67 if (pt>=0) G.pb(MP(F[ls][i].fi,mn[pt])); 68 } 69 return; 70 } 71 inline void dfs(ll u,ll fa) 72 { 73 int lson=0,rson=0; 74 F[u].clear(); 75 if (du[u]==1){F[u].pb(MP(0,0));return;} 76 GO(u) 77 { 78 ll v=e[j].v; 79 if (v==fa) continue; 80 if (!lson) lson=v; 81 else rson=v; 82 dfs(v,u); 83 } 84 GO(u) 85 { 86 ll v=e[j].v; 87 if (v==fa) continue; 88 FOR(i,0,(ll)F[v].size()-1) 89 { 90 F[v][i].fi+=e[j].w; 91 F[v][i].se+=e[j].w; 92 } 93 } 94 update(F[u],lson,rson); 95 return; 96 } 97 int main() 98 { 99 // freopen("data.in","r",stdin); 100 mem(f,-1); 101 n=read(); 102 FOR(i,2,n) a[i]=read(),v[i]=read(),add(i,a[i],v[i]),add(a[i],i,v[i]),du[i]++,du[a[i]]++; 103 L=0,R=1e9; 104 while (L<R) 105 { 106 MID=(L+R)>>1; 107 dfs(1,0); 108 if (F[1].size()) R=MID; 109 else L=MID+1; 110 } 111 write(L); 112 return 0; 113 }
T3 Mole and Abandoned Mine
Effect: Gives a 15 point undirected weighted graph, ask how Erase takes a minimum cost, from 1 to n such that there is a unique simple path
answer:
- If we know that the final piece of the path from 1 to n, then the two adjacent points must be on the side of the bridge connection path, i.e. each following points can be linked to a communication block
- DP design condition [i] [S] represents the path from 1 to i, where S arranged so many points, right up to the edge of reservations
转移：dp [ j ][ T ] = max{ dp [ i ] [ S ] + wi,j + val [ T ^ S ] }
1 #include<bits/stdc++.h> 2 #define FOR(i,a,b) for (register int i=(a);i<=(b);i++) 3 #define For(i,a,b) for (register int i=(a);i>=(b);i--) 4 #define mem(i,j) memset(i,j,sizeof(i)) 5 #define GO(u) for (register int j=f[u];j!=-1;j=nxt[j]) 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define pii pair<int,int> 10 #define MP make_pair 11 using namespace std; 12 typedef long long ll; 13 const int N=20; 14 const int M=500; 15 const int P=1<<16; 16 int n,m,x,y,z,dp[N][P],G[N][N],tot,sum=0,g[P],ans; 17 inline int read() 18 { 19 int x=0,f=1; 20 char c=getchar(); 21 while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} 22 while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();} 23 return f*x; 24 } 25 inline void write(int x) 26 { 27 if (x<0) putchar('-'),x=-x; 28 if (x>9) write(x/10); 29 putchar(x%10+'0'); 30 return; 31 } 32 int main() 33 { 34 // freopen("data.in","r",stdin); 35 mem(G,-1); 36 n=read(),m=read(); 37 FOR(i,1,m) 38 { 39 x=read(),y=read(),z=read(); 40 sum+=z; 41 G[x][y]=G[y][x]=z; 42 } 43 tot=(1<<n)-1; 44 FOR(i,1,tot) 45 FOR(j,0,n-1) if ((i>>j)&1) 46 FOR(k,j,n-1) if ((i>>k)&1) 47 if (G[j+1][k+1]!=-1) g[i]+=G[j+1][k+1]; 48 mem(dp,-1); 49 dp[1][1]=0; 50 FOR(S,1,tot) 51 { 52 FOR(i,0,n-1) if (((S>>i)&1)&&dp[i+1][S]!=-1) 53 { 54 int ss=(tot xor S)|(1<<i); 55 for (register int s=ss;s;s=(s-1)&ss) 56 { 57 if (!((s>>i)&1)) continue; 58 dp[i+1][S|s]=max(dp[i+1][S|s],dp[i+1][S]+g[s]); 59 } 60 FOR(j,0,n-1) 61 { 62 if ((S>>j)&1) continue; 63 if (G[i+1][j+1]==-1) continue; 64 dp[j+1][S|(1<<j)]=max(dp[j+1][S|(1<<j)],dp[i+1][S]+G[i+1][j+1]); 65 } 66 } 67 } 68 ans=sum-dp[n][tot]; 69 write(ans); 70 return 0; 71 }