51nod1253 Kundu and Tree

The tree consists of n points and N-1 edges. The sides of the tree have 2 colors red (' r ') and Black (' B '). Given the color of the N-1 edges, ask how many triples (a,b,c) of nodes are satisfied: Node A to Node B, node B to node C, node c to Node A, each path has at least one edge red. Note (a,b,c), (B,a,c), and all other permutations are considered to be the same ternary group. The result of the output is 1000000007.   Input

Line 1th: 1 number N (1 <= n <= 50000) line 2-n: 2 numbers per line plus one color, indicating the start and end of edges and color.

Output

Outputs 1 numbers, corresponding to the number of 3 tuples that meet the criteria.

preprocessing for each edge (A, A, b), the path from a to go through the number of paths is a/no red edge

There are two types of situations:

1.a,b,c 22 paths do not go through the 3rd, when three paths have a unique common point, the common point to A,b,c path at least two have red edge, thus can be counted

A path with two points in the 2.a,b,c is passed through the 3rd, when the enumeration is counted.

#include <cstdio>typedefLong Longi64;Const intn=50007, r=n* +;Charbuf[r+7],*ptr=buf-1;int _(){    intx=0, c=*++ptr;  while(c< -) c=*++ptr;  while(c> -) x=x*Ten+c- -, c=*++ptr; returnx;}int_c () {intc=*++ptr;  while(c<'a') c=*++ptr; returnc=='R';}intN;intes[n*2],enx[n*2],e0[n],ev[n*2],ep=2;intf1[n],f2[n],f3[n],sz[n];i64 ans=0;voidDFS1 (intWintPA) {Sz[w]=1;  for(intI=e0[w];i;i=Enx[i]) {        intu=Es[i]; if(u!=PA)            {DFS1 (u,w); SZ[W]+=Sz[u]; if(Ev[i]) f1[w]+=f3[u]=Sz[u]; Elsef1[w]+=f3[u]=F1[u]; }    }}voidDFS2 (intWintPA) {     for(intI=e0[w];i;i=Enx[i]) {        intu=Es[i]; if(u!=PA) {            if(Ev[i]) f2[u]=n-Sz[u]; Elsef2[u]=f2[w]+f1[w]-F1[u];        DFS2 (U,W); }    }}voidDFS3 (intWintPA) {    Statici64 Fs[n],fl[n],fr[n],ss[n],sl[n],sr[n]; Static intp;  for(intI=e0[w];i;i=Enx[i]) {        intu=Es[i]; if(u!=pa) dfs3 (u,w); } P=0;  for(intI=e0[w];i;i=Enx[i]) {        intu=Es[i]; ++p; if(u!=PA) {Fs[p]=fl[p]=fr[p]=F3[u]; SS[P]=sl[p]=sr[p]=sz[u]-Fs[p]; }Else{Fs[p]=fl[p]=fr[p]=F2[w]; SS[P]=sl[p]=sr[p]=n-sz[w]-Fs[p]; }} i64 A0=ans;  for(intI=2; i<=p;++i) fl[i]+=fl[i-1],sl[i]+=sl[i-1];  for(inti=p-1; i;--i) fr[i]+=fr[i+1],sr[i]+=sr[i+1];  for(intI=2; i<p;++i) {ans+=fl[i-1]* (fs[i]*sr[i+1]+ss[i]*fr[i+1]); Ans+ = (sl[i-1]+fl[i-1]) *fs[i]*fr[i+1]; }     for(intI=1; i<p;++i) ans+=fl[i]*fs[i+1];}intMain () {fread (buf,1, R,stdin); N=_();  for(intI=1; i<n;++i) {        intA=_ (), B=_ (), c=_c (); ES[EP]=b;enx[ep]=e0[a];ev[ep]=c;e0[a]=ep++; ES[EP]=a;enx[ep]=e0[b];ev[ep]=c;e0[b]=ep++; } DFS1 (1,0); DFS2 (1,0); DFS3 (1,0); printf ("%lld", ans%1000000007); return 0;}

51nod1253 Kundu and Tree