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水题一个!308K 0ms 数据弱,不需要过多纠结考慮精度問題#include <iostream>
#include <stdio.h>
using namespace std;
int N;
double h[35][35];
double dp[35][35][35];
int trc[35][35][35];
int main() {
while(scanf("%d", &N) > 0){
for(int i = 1; i <= N; i++){
for(int j = 1; j <= N; j++){
if(i == j) h[i][j] = 0.0;
else scanf("%lf", &h[i][j]);
}
}
for(int i = 1; i <= N; i++){
for(int j = 1; j <= N; j++){
//if(i == j) continue;
dp[1][i][j] = h[i][j];
trc[1][i][j] = i;
}
}
int startP = -1, cs = 2333;
for(int k = 2; k <= N; k++){
for(int i = 1; i <= N; i++){
for(int j = 1; j <= N; j++){
double mx = 0.0;
int arg = -1;
for(int j_ = 1; j_ <= N; j_++){
if(j_ == j) continue;
double temp = dp[k-1][i][j_] * h[j_][j];
if(temp > mx){
mx = temp;
arg = j_;
}
}
dp[k][i][j] = mx;
trc[k][i][j] = arg;
if(i == j && mx > 1.01){
startP = i;
cs = k;
goto done;
}
}
}
}
done:
if(startP == -1){
printf("no arbitrage sequence exists\n");
continue;
}
int seq[35];
seq[cs] = startP;
int pos = cs;
int cur = trc[cs][startP][startP];
while(pos > 0){
pos--;
seq[pos] = cur;
cur = trc[pos][startP][cur];
}
for(int i = 0; i <= cs; i++){
printf(" %d", seq[i]);
}
printf("\n");
}
return 0;
}
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