题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1260
分析：先算一下数据范围最多50000s，所以不超过当天晚上12点；
dp相当水，从前往后决策
dp[i] = min(dp[i - 1] + s[i], dp[i - 2] + d[i])
1 #include<iostream> 2 #include<sstream> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<string> 6 #include<cstring> 7 #include<algorithm> 8 #include<functional> 9 #include<iomanip> 10 #include<numeric> 11 #include<cmath> 12 #include<queue> 13 #include<vector> 14 #include<set> 15 #include<cctype> 16 #define PI acos(-1.0) 17 const int INF = 0x3f3f3f3f; 18 const int NINF = -INF - 1; 19 const int maxn = 2e3 + 5; 20 typedef long long ll; 21 using namespace std; 22 int s[maxn], d[maxn]; 23 int dp[maxn]; 24 int main() 25 { 26 int n; 27 scanf("%d", &n); 28 while (n--) 29 { 30 int k; 31 scanf("%d", &k); 32 for (int i = 0; i <= k; ++i) 33 s[i] = INF, d[i] = INF, dp[i] = 0; 34 for (int i = 1; i <= k; ++i) 35 scanf("%d", &s[i]); 36 for (int i = 2; i <= k; ++i) 37 scanf("%d", &d[i]); 38 dp[1] = s[1]; 39 for (int i = 2; i <= k; ++i) 40 { 41 dp[i] = min(dp[i - 1] + s[i], dp[i - 2] + d[i]); 42 //cout << i << ' ' << dp[i] << endl; 43 } 44 int h, m, s; 45 h = 8 + dp[k] / 3600; 46 m = (dp[k] % 3600) / 60; 47 s = (dp[k] % 3600) % 60; 48 printf("%02d:%02d:%02d ", h, m, s); 49 if (h >= 12) printf("pm "); 50 else printf("am "); 51 } 52 return 0; 53 }