bzoj1063: [Noi2008] Road design

Tree-shaped DP.

This problem is divided into the same way as the tree chain, but the heavy path can pass through the root node, and requires the number of schemes, so can not be used to split the tree chain.

But by this relationship you can tell that the answer is very small for the log n level (after looking at the data, the answer is the maximum of 4). But there should be more data to make the answer bigger.

With F[I][J][0/1/2], respectively, in the node with I as a subtree, the point of the most convenient value is J.

0 means that the I node is not in the chain of any one son, 1 means that in a chain composed of a son, 2 represents a chain of 2 sons that passes through the I node.

(This is a hard to understand feeling.) ）

A recursive relationship between them ... Then I won't be typing under this box, as explained above.

F1 calculates the number of scenarios that v[i] can link to the root node. F2 represents the number of scenarios that are not connected to the root node.

 f1=f[v[i]][j][0 ]+f[v[i]][j][1  Span style= "COLOR: #000000" >]f2  = (j==0 ?  0 :(F[v[i]][j-1 ][0 ]+f[v[i]][j-< Span style= "COLOR: #800080" >1 ][1 ]+f[v[i]][j- 1 ] [2   2 ]=f[u][j][1 ]*f1+f[u][j ][2 ]*f2;f[u][j][ 1 ] =f[u][j][0 ]*f1+f[u][j][1 ]*f2;f[u][j][ 0 ]=f[u][j][0 ]*f2; 

#include <cstdio>#include<algorithm>#include<cstring>#defineLL Long Longusing namespacestd;Const intMAXN =200000+Ten;Const intMaxh = A;intG[maxn],v[maxn],next[maxn],eid;Long Longf1,f2,f[maxn][maxh][3],res,mod;intN,m,ans;voidAddedge (intAintb) {V[eid]=b; Next[eid]=g[a]; g[a]=eid++; V[eid]=a; NEXT[EID]=G[B]; g[b]=eid++;} ll update (ll x) {if(X%mod)returnX%MoD; returnx==0?0: mod; }voidDfsintUintFA) {     for(intI=g[u];~i;i=next[i])if(v[i]!=FA)        {DFS (V[I],U);  for(intj=0; j<maxh;j++) {F1=update (f[v[i]][j][0]+f[v[i]][j][1]); F2= (j==0?0: Update (f[v[i]][j-1][0]+f[v[i]][j-1][1]+f[v[i]][j-1][2])); f[u][j][2]=update (f[u][j][1]*f1+f[u][j][2]*F2); f[u][j][1]=update (f[u][j][0]*f1+f[u][j][1]*F2); f[u][j][0]=update (f[u][j][0]*F2); }    }}intMain () {memset (g,-1,sizeof(g)); scanf ("%d%d%lld",&n,&m,&MoD); if(m!=n-1) {printf ("-1\n-1\n"); return 0; }     for(intI=1, u,v;i<n;i++) {scanf ("%d%d",&u,&v);    Addedge (U,V); }     for(intI=1; i<=n;i++)     for(intj=0; j<maxh;j++) f[i][j][0]=1; DFS (1,0);  for(intI=0, res;i<maxh;i++) {res=update (f[1][i][0]+f[1][i][1]+f[1][i][2]); if(res) {ans=res%MoD; printf ("%d\n%d\n", I,ans);  Break; }    }    return 0;}

bzoj1063: [Noi2008] Road design