Idea: For a tree, for any node, its maximum value can come from a subtree rooted in it, or it can come from its father.
It is therefore necessary to consider the situation in two directions. Therefore, 2 directions must be considered.
Solution: Dp[i][0] Save subtree value, Dp[i][1] Save Father node Direction +dis[i,root];
You can save the maximum value of any subtree by first DFS, while preserving the direction of the maximum at any point while maintaining his second largest value.
Then the second time Dfs, for node u, if its child is not it to take the maximum path on the point, then the child's value dp[i][1] is
Max (dp[root][1], dp[root][0]) + dis[i,root]; Dp[i][1] is the second largest value of the Father node + dis[i,root];
Then the second largest value is dp[i][1];
#include <cstdio>#include<cstring>#include<iostream>using namespacestd;Const intMAXN =10010;structnode{intto ; intv; intNext;} EDGE[MAXN*3];intPre[maxn],index,vis[maxn],n;intdp[maxn][2],WAY[MAXN],SV[MAXN];//sv[] On the other side of the way[that is not a subtree] indicates where the maximum value of the current point comes from//Dp[][0] Indicates the maximum value of the subtree direction dp[][1] represents the maximum value in the other directionintMaxintXinty) { returnX>y?x:y;}intMinintXinty) { returnX<y?x:y;}voidAddintXintYintz) {edge[index].to=y; EDGE[INDEX].V=Z; Edge[index].next=Pre[x]; PRE[X]= index++;}voidDFS1 (introot) {Vis[root]=1; inti; int Set=Root; for(I=pre[root]; i!=-1; I=Edge[i].next) { intt =edge[i].to; if(!Vis[t]) {DFS1 (t); if(dp[root][0] > dp[t][0] +edge[i].v) {Sv[root]= Max (Sv[root], dp[t][0] +edge[i].v); } Else { Set=T; Sv[root]= dp[root][0]; dp[root][0] = dp[t][0] +edge[i].v; }}} Way[root]=Set;}voidDFS2 (introot) {Vis[root]=1; inti; for(I=pre[root]; i!=-1; I=Edge[i].next) { intt =edge[i].to; if(!Vis[t]) { if(Way[root]! =t) {dp[t][1] = max (dp[root][0], dp[root][1]) +edge[i].v; Sv[t]= dp[t][1]; } Else{dp[t][1] = Sv[root] +edge[i].v; Sv[t]= dp[t][1]; } DFS2 (t); } }}intMain () {inti; while(~SCANF ("%d",&N)) {index=1; memset (SV,0,sizeof(SV)); Memset (DP,0,sizeof(DP)); memset (PRE,-1,sizeof(pre)); for(i=2; i<=n; ++i) { intx,v; scanf ("%d%d", &x, &v); Add (i, X, V); Add (x, I, v); } memset (Vis,0,sizeof(VIS)); DFS1 (1); /*for (i=1; i<=n; i++) {printf ("%d", sv[i]); } cout<<endl;*/memset (Vis,0,sizeof(VIS)); DFS2 (1); for(i=1; i<=n; i++) {printf ("%d\n", Max (dp[i][0],dp[i][1])); } }}/*1 of 181 , one by one*/
hdu2196 Tree DP