題意:給出了n種貨币,然後給出了每種貨币和其他n-1種貨币的匯率,問是否能有一種貨币換算情況使一種貨币能從本身1經過換算後增加0.01以上,如果有輸出最短的換算方式,否則輸出no。
題解:dp題,用f[t][i][j]儲存在第t次從貨币i換算到貨币j的最大換算出的錢,狀态轉移方程就是f[t][i][j] = max{f[t][i][j], f[t - 1][i][k] * m[k][j] },其中k是枚舉的中轉點,如果f[t][i][i] > 1.01說明成功,然後用path[t][i][j]存儲第t次換算從貨币i到j的中轉點,遞歸列印路徑。
#include <stdio.h>
#include <string.h>
const int N = 25;
int n;
double m[N][N];
double f[N][N][N];
int path[N][N][N];
void print(int t, int i, int j) {
if (t == 1) {
printf("%d", i);
return;
}
print(t - 1, i, path[t][i][j]);
printf(" %d", path[t][i][j]);
}
bool dp() {
for (int t = 2; t <= n; t++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
double maxx = -1;
int index;
for (int k = 1; k <= n; k++)
if (maxx < f[t - 1][i][k] * m[k][j]) {
maxx = f[t - 1][i][k] * m[k][j];
index = k;
}
f[t][i][j] = maxx;
path[t][i][j] = index;
}
if (f[t][i][i] - 1.01 > 1e-9) {
print(t, i, i);
printf(" %d\n", i);
return true;
}
}
}
return false;
}
int main() {
while (scanf("%d", &n) != EOF) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (i != j) {
scanf("%lf", &m[i][j]);
f[1][i][j] = m[i][j];
}
int res = dp();
if (!res)
printf("no arbitrage sequence exists\n");
}
return 0;
}