链接
传送门
题意
有\(n\)个城市，要从第1个出发，经过\(k\)天到第\(n\)个。给出每天城市间旅行的花费，输出最小的话费值。
思路
定义\(dp[d][i]\)为第\(d\)天到城市\(i\)的最小花费。如果第\(d + 1\)天在城市\(i\)与城市\(j\)之间有航线，则可以转移至\(dp[d + 1][j]\)。
代码
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| #include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 12;
const int maxd = 35;
const int maxk = 1010;
int c[maxn][maxn][maxd], num[maxn][maxn];
int dp[maxk][maxn];
int main() {
freopen("/Users/wangchengrui/Desktop/in.txt", "r", stdin);
int n, k, t = 0;
while (~scanf("%d%d", &n, &k) && (n || k)) {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i != j) {
scanf("%d", &num[i][j]);
for (int d = 0; d < num[i][j]; ++d) {
scanf("%d", &c[i][j][d]);
}
}
}
}
memset(dp, 0x3f, sizeof dp);
dp[0][1] = 0;
for (int d = 1; d <= k; ++d) {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i != j) {
int x = (d - 1) % num[i][j];
if (c[i][j][x] != 0) {
dp[d][j] = min(dp[d][j], dp[d - 1][i] + c[i][j][x]);
}
}
}
}
}
printf("Scenario #%d\n", ++t);
if (dp[k][n] != inf) {
printf("The best flight costs %d.\n\n", dp[k][n]);
} else {
puts("No flight possible.\n");
}
}
return 0;
}
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