error 3259

the time tunnel)#include #include#includeusing namespacestd;Const intn=510;Const intinf=0x3f3f3f3f;intG[n][n], Dist[n], cou[n], vis[n], N;///cou Array saves the number of times the point is traversedvoidInit () {intI, J; for(i =1; I ) {Dist[i]=INF; Cou[i]= Vis[i] =0; for(j =1; J ) G[i][j]=INF; G[i][i]=0; }}intSPFA () {intI, V; Queueint>p; Q.push (1); dist[1] =0; cou[1]++; while(!Q.empty ()) {v=Q.front (); Q.pop (); for(i =1; I ) { if(Dist[i] > g[v][i]+Dist[v]) {Dist

, T) that describe, respectively: A bidirectional path between S and E that requires T seconds to traverse. The might is connected by more than one path.Lines M+2.. M+W+1 of each farm:three space-separated numbers (S, E, T) that D Escribe, respectively:a One, path from S to E . Also moves the traveler back T sec Onds.OutputLines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (does not include the quotes).Sample Input23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43

(￣▽￣) " #include #include#include#include#include#includeusing namespacestd;Const intinf=10e7;Const intmaxn=510;intf,n,m,w,s,e,t;intLC[MAXN],CNTNODE[MAXN];BOOLVIS[MAXN];structedge{intV,cost;} E;vectorEDGE[MAXN];BOOLSPFA () {Queueint>que; intU,v,c,len; for(intI=1; i) Lc[i]=inf,vis[i]=0, cntnode[i]=0; lc[1]=0; vis[1]=1; cntnode[1]=1; Que.push (1); while(!Que.empty ()) {u=Que.front (); Que.pop (); Vis[u]=0; len=edge[u].size (); for(intI=0; i) {v=edge[u][i].v; C=Edge[u][i].cost; if(lc[v]>lc[u]+c)

(); + Que.pop (); AVis[temp] =false; the for(inti = head[temp]; ~i; i =Edge[i].next) { + intv =edge[i].to; - if(D[v] > D[temp] +edge[i].cost) { $D[V] = D[temp] +Edge[i].cost; $ if(!Vis[v]) { - Que.push (v); -VIS[V] =true; thecnt[v]++; - if(Cnt[v] >=N)Wuyi return true; the } - } Wu } - } About return false; $ } - - intMain () - { A intN, M, C, U, V, cost,

integers respectively:N,M, andWLines 2..M+1 of each farm:three space-separated numbers (S,E,T) that describe, respectively:a bidirectional path betweenSandEThat requiresTSeconds to traverse. The might is connected by more than one path.LinesM+2..M+W+1 of each farm:three space-separated numbers (S,E,T) that describe, respectively:a one-path fromSToEThat also moves the traveler backTSeconds.OutputLines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (does not i

]; - //For (i = 0; i Amemset (Arr_d,0,sizeof(int) *n); + the //Bellman-ford - BOOLFlag; $ for(i =0; I 1; ++i) the { theFlag =false; the for(j =0; J 2* m + W; ++j) the { - if(ARR_D[ARR_ROAD[J].E] > ARR_D[ARR_ROAD[J].S] +arr_road[j].t) in { theARR_D[ARR_ROAD[J].E] = Arr_d[arr_road[j].s] +arr_road[j].t; theFlag =true; About } the } the if(!flag) Break; the

integers respectively:N,M, andWLines 2..M+1 of each farm:three space-separated numbers (S,E,T) that describe, respectively:a bidirectional path betweenSandEThat requiresTSeconds to traverse. The might is connected by more than one path.LinesM+2..M+W+1 of each farm:three space-separated numbers (S,E,T) that describe, respectively:a one-path fromSToEThat also moves the traveler backTSeconds.OutputLines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (does not i

]++; - while(!q.empty ()) - { in intu=Q.front (); - Q.pop (); tovis[u]=0; + for(intI=1; i) - { the if(d[i]>d[u]+G[u][i]) * { $d[i]=d[u]+G[u][i];Panax Notoginseng if(!Vis[i]) - { the if(used[i]>=MAXN) + return true; A Q.push (i); thevis[i]=1; +used[i]++; - } $ } $ } - } - return false; the } - Wuyi intMain () the { - intt

The experience of having a previous problem with a template problem is easy.Except because SPFA is not very familiar with the wrong two-letter and then debug a little#include #includestring>#include#include#include#include#include#include#includeSet>#include#defineINF 0x3f3f3f3fusing namespacestd;intSize, head[1010], next[6010], point[6010], val[6010];voidinit () {size=0; memset (Head,-1,sizeofhead);}voidAddint from,intTo,intvalue) {Point[size]=to ; Val[size]=value; Next[size]= head[ from]; hea

bidirectional path betweenSandEThat requiresTSeconds to traverse. The might is connected by more than one path.LinesM+2..M+W+1 of each farm:three space-separated numbers (S,E,T) that describe, respectively:a one-path fromSToEThat also moves the traveler backTSeconds.OutputLines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (does not include the quotes).Sample Input23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8Sample OutputNOYES/* Topic: Wormhole problem, no

#define _crt_secure_no_warnings#includeIf you use STL to make adjacency matrix, it will time out, the memory request release cost is too large. and the title POJ 3259 wormholes AC code (negative weight ring judgment, Bellmanford)

; + } - $ voidAddedges (intAintBintc) { $e[++ecnt].u=A; -e[ecnt].v=b; -e[ecnt].cost=C; the } - Wuyi intMain () { the intm,w,ncase; - inta,b,c; Wuscanf"%d",ncase); - while(ncase--){ AboutEcnt=0; $scanf" %d%d%d",nodenum,m,W); - while(m--){ -scanf" %d%d%d",a,b,c); - addedges (a,b,c); A addedges (B,A,C); + } the while(w--){ -scanf" %d%d%d",a,b,c); $c=-C; the addedges (a,b,c); the } the if(Bellman_ford ()) printf ("no\n"); the Elsepri

Http://poj.org/problem? Id = 3259 Use Bellman-Ford and spfa to determine negative loops... Spfa [856kb 125ms] # Include Bellman-Ford [736kb 125ms] # Include

is good at primary school physics ~ ~ ~), each wormhole has a backward time, he would like to use these wormhole back to the past. Ask him if he can achieve this magical journey.Analysis: Is the simple judgment whether there is a negative ring, because if there is a negative ring, it represents the return to the origin of the time before the departure, this will be before. But pay special attention to the road is two-way, but the wormhole is one-way!!! Level of the array must be open enough, or

Wormholes Time limit:2000ms Memory limit:65536k Total submissions:35456 accepted:12955 Description While exploring he many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it's a one-way path

Description While processing his own farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is before you entered the wormhole!

This question can also be solved using the spfa algorithm. However, the key point is that when the number of columns of a vertex exceeds the total number of all vertices, it can be determined that a negative ring exists. Spfa can also process

Wormholes Time limit:2000 ms   Memory limit:65536 K Total submissions:30766   Accepted:11157 Description While processing his own farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very

Wormholes Time Limit:2000 MS   Memory Limit:65536 K Total Submissions:25394   Accepted:9083 Description While processing his own farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very

This question is very interesting. Ask if the question can be made between time and space. This is very similar to the previous poj1860, but it only describes two points: 1. the routing is bidirectional, and the holes are unidirectional. 2. In the