Arbitrage POJ 2240

http://poj.org/problem?id=2240

Test instructions: Now you have n kinds of currency, after the currency is exchanged, eventually still need to be converted into the original currency, if one of the currencies increased, output "Yes", otherwise output "No".

At first, tle, because the position of return 1 was misplaced. At first, I did not think, directly to the final judgment ...

#include <iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstdlib>#include<cstring>#include<map>#include<queue>using namespacestd;#defineMAXN 150#defineINF 0x3f3f3f3fintHEAD[MAXN], V[MAXN];DoubleDIST[MAXN];intCNT;structnode{intu, V, next; DoubleW;} MAPS[MAXN];voidADD (intUintVDoubleW) {MAPS[CNT].V=v; MAPS[CNT].W=W; Maps[cnt].next=Head[u]; Head[u]= cnt + +;}intSPFA (ints) {memset (V,0,sizeof(v)); Queue<int>Q;    Q.push (s); V[s]=1;  while(Q.size ()) {intp=Q.front ();        Q.pop (); V[P]=0;  for(intI=HEAD[P]; i!=-1; I=Maps[i].next) {            intQ =maps[i].v; Doublet = dist[p]*MAPS[I].W; if(dist[q]<t) {Dist[q]=T; if(!V[q]) {V[q]=1;                Q.push (q); }            }        }        if(Dist[s] >1)        return 1; }    return 0;}intMain () {intN, M, t=1; Charstr[ -];  while(SCANF ("%d", &N), N) {map<string,int>s;        S.clear ();  for(intI=1; i<=n; i++) {scanf ("%s", str); S[STR]=i; } scanf ("%d", &m); Chara[ -], b[ -]; DoubleC; CNT=0; memset (Head,-1,sizeof(head));  for(intI=1; i<=m; i++) {scanf ("%s%lf%s", A, &C, B);        ADD (S[a], s[b], c); }        intAns =0;  for(intI=1; i<=n; i++)        {             for(intj=1; j<=n; J + +) Dist[j]= -INF; Dist[i]=1; Ans+=SPFA (i); if(ANS) Break; }        if(ANS) printf ("Case %d:yes\n", t++); Elseprintf"Case %d:no\n", t++); }    return 0;}

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Arbitrage POJ 2240