Ultraviolet A 104-arbitrage

Source: Internet
Author: User

// Start to use the Implicit Graph search, and the results exceed the memory. Remember that the time is not calculated before you do the question, and the memory limit is faulty.

// First, it indicates that this question can use dynamic rules. The difficulty lies in how to output the shortest switching mode. In this light, one dimension is required to mark [STEP].

// Then Floyd + motion gauge, O (n4) is enough




# Include <iostream> <br/> # include <memory. h> <br/> # include <cstdio> <br/> # include <stack> </P> <p> using namespace STD; <br/> const int max = 25; <br/> const Double E = 1e-9, Sta = 1.01; <br/> bool findflag; <br/> int N, sign, totalstep; <br/> double map [Max] [Max] [Max], way [Max] [Max]; <br/> stack <int> S; <br/> inline int dbcmp (const double & X) {<br/> If (x>-E & x <E) <br/> return 0; <br/> return x> 0? 1:-1; <br/>}< br/> int main () <br/>{< br/> freopen ("I .txt", "r", stdin ); <br/> int K; <br/> while (CIN> N) {<br/> memset (way, 0, sizeof (way )); <br/> memset (MAP, 0, sizeof (MAP); <br/> for (INT I = 1; I <= N; I ++) {<br/> for (Int J = 1; j <= N; j ++) {<br/> if (I = J) {<br/> map [I] [J] [1] = 1.0; <br/> way [I] [J] [1] = I; <br/> continue; <br/>}< br/> CIN> map [I] [J] [1]; <br/> way [I] [J] [1] = I; <br/>}< br/> findflag = 0; <br/> For (INT step = 2; Step <= N; Step ++) {<br/> for (k = 1; k <= N; k ++) {<br/> for (INT I = 1; I <= N; I ++) {<br/> for (Int J = 1; j <= N; j ++) {<br/> If (Map [I] [J] [STEP] <map [I] [k] [step-1] * map [k] [J] [1 ]) {<br/> map [I] [J] [STEP] = map [I] [k] [step-1] * map [k] [J] [1]; <br/> way [I] [J] [STEP] = K; <br/>}< br/> if (I = J) <br/> If (dbcmp (Map [I] [J] [STEP]-Sta) = 1) {<br/> findflag = 1; <br/> Sign = J; <br/> totalstep = step; <br/> goto ans; <br/>}< br/>} <Br/>}< br/> ans: <br/> If (findflag) {<br/> while (! S. empty () s. pop (); <br/> int end = sign; <br/> S. push (sign); <br/> for (INT I = totalstep; I> = 2; end = way [sign] [end] [I], I --) <br/> S. push (way [sign] [end] [I]); <br/> cout <way [sign] [end] [1]; <br/> while (! S. empty () {<br/> cout <"" <S. top (); <br/> S. pop (); <br/>}< br/> cout <Endl; <br/>}< br/> else <br/> cout <"no arbitrage sequence exists" <Endl; <br/>/* For (INT I = 1; I <= N; I ++) {<br/> for (Int J = 1; j <= N; j ++) {<br/> cout <map [I] [J] <""; <br/>}< br/> cout <Endl; <br/>}*/<br/>}< br/> return 0; <br/>}</P> <p>

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