tree silhouette free

++; - if(!road[i].l) About { $road[i].bj=1; -js2++; - } - HB (x, y); A } + if(js1==n-1) Break; the } - if(js1!=n-1|| Js2>t) $ { theprintf"No solution\n"); theExit0); the } theSort (road+1, road+1+m,px2); -js1=0, js2=0; in for(intI=1; ii; the for(intI=1; i) the { About if(ROAD[I].BJ) the { the intX=road[i]. from, y=road[i].to; the HB (x, y); +js2++; -js1++; theans[js1]=i;Bayi } th

3624: [Apio2008] Free road time limit:2 Sec Memory limit:128 mbsec Special Judgesubmit:111 solved:49[Submit] [Status] DescriptionInputOutputSample Input5 7 21 3 04 5 13 2 05 3 14 3 01 2 14 2 1Sample Output3 2 04 3 05 3 11 2 1 or look at the online standard ..... The practice of this problem is first 1 side priority, do the spanning tree, if the number of 0 sides is greater than k no solution, or if

This is the first of a series of Martian previews (one, two, three, four, five questions and answers, six, seven). One: Demand and story structure Second: Edit stories, product management, organizational structure The third: iteration, planning, assigning tasks Part Four: Storyboard, burndown chart, my work item V: FAQ The six: My Space, my notice The Seven: custom fields In addition to doing training, blogging, editing manuals, two years to use half an amateur half full-time time 1.

=Find (v); if(x==y)Continue; FA[X]=y; Ans[++p]=b[i]; flag[i]=1; if(++cnt = = N1) Break; } if(P > K | | cnt 1) {puts ("No solution");return 0;} for(intI=1; ii; for(intI=1; ifind (ANS[I].V); CNT=p; if(cnt for(intI=1; iif(!Flag[i]) {u=B[I].U, v=b[i].v; intX=find (u), y=Find (v); if(x==y)Continue; FA[X]=y; ans[++cnt]=B[i]; if(cnt = = k) Break; } for(intI=1; i) {u=A[I].U, v=a[i].v; intX=find (u), y=Find (v); if(x==y)Continue; FA[X]=y; ans[++cnt]=A[i]; if(cnt = = N1) Break; } for(intI=1;

Free Pies bzoj-2131Main Topic :Note : $1\le n \le 10^5$,$1\le w \le 10^8$.idea : First, think of DPstatus : Dp[i][j] represents the maximum benefit of I minute in position JOptimized optimizationstatus : Dp[i] Indicates the maximum benefit of the last received I.Transfer : Order enumeration i:1->n.then , we try to optimizeFor this state we will find that there is an absolute dead thing at the time of transfer, and we take it apart:2*t[j]+pos[j]Then, b

; GETG (V,U,DEP+COLOR[V],VAL+EDGE[I].W); }}voidWorkintu) {vis[u]=1;intI, J; for(i=head[u];i!=-1; i=edge[i].next) {intV=EDGE[I].V;if(Vis[v])Continue; GetSize (v,-1); Min1=inf; Getroot (v,-1, Siz[v]); Work (root); }inttot=0; for(i=head[u];i!=-1; i=edge[i].next) {intV=EDGE[I].V;if(Vis[v])Continue; Getnum (v,-1); T[tot].v=v; T[TOT].NUM=NUM[V]; T[TOT].W=EDGE[I].W; tot++; } sort (t,t+tot,cmp);intLim=k-color[u]; for(i=0; i

; thetypedef vectorVL; thetypedef vectorVVL; thetypedef vectorBOOL>vb; - in Const intMAXN =100100; the ConstLL mod = 1e9+7; the intN, M; About intA[MAXN], B[MAXN]; the VI G[MAXN]; the BOOLVIS[MAXN]; the LL DP[MAXN]; + LL ret; - the voidDfsintu) {Bayi Rep (i, g[u].size ()) { the intv =G[u][i]; the if(!Vis[v]) { -VIS[V] =1; - Dfs (v); the } theDp[u] = (Dp[u] + dp[v])%MoD; the } theRET = (ret + (dp[u] * a[u]% MoD))%MoD; -Dp[u] = (Dp[u] + b[u])%MoD; the } the the intMa