暑假集训-合训第三场

Origin	Title
31 / 51	Problem A	HDU 4175	Class Schedule
27 / 140	Problem B	HDU 4193	Non-negative Partial Sums
5 / 17	Problem C	HDU 4212	Shut the Box
3 / 9	Problem D	HDU 4216	Computational Geometry?
2 / 3	Problem E	HDU 4219	Randomization?
	Problem F	HDU 4234	Moving Points
9 / 17	Problem G	HDU 4243	Maximum in the Cycle of 1
	Problem H	HDU 4253	Two Famous Companies
5 / 25	Problem I	HDU 4254	A Famous Game
31 / 141	Problem J	HDU 4268	Alice and Bob
23 / 46	Problem K	HDU 4260	The End of The World
56 / 167	Problem L	CodeForces 557B	Pasha and Tea

　　　　题意：有C类课程，每类课程有T个班，不同的班在不同位置，上不同的班有不同的花费，初始位置是0，末位置L，求最小花费。

　　　　比较简单的递推，f[i][j]表示上i类课的j个班所需的最小花费。最后求f[c]中的最小值就可以了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define eps 1e-8
16 #define INF 0x3f3f3f3f
17 //#define OPEN_FILE
18 using namespace std;
19 int T, c, t, L;
20 struct Node{
21     int pos, v;
22 }p[30][1005];
23 int f[30][1005];
24 int main()
25 {
26 #ifdef OPEN_FILE
27     freopen("in.txt", "r", stdin);
28     //freopen("out.txt", "w", stdout);
29 #endif // OPEN_FILE
30     scanf("%d", &T);
31     for (int cas = 1; cas <= T; cas++){
32         scanf("%d%d%d", &c, &t, &L);
33         for (int i = 1; i <= c; i++){
34             for (int j = 1; j <= t; j++){
35                 scanf("%d%d", &p[i][j].pos, &p[i][j].v);
36             }
37         }
38         memset(f, INF, sizeof(f));
39         for (int i = 1; i <= t; i++){
40             f[1][i] = p[1][i].pos + p[1][i].v;
41         }
42         for (int i = 2; i <= c; i++){
43             for (int j = 1; j <= t; j++){
44                 for (int k = 1; k <= t; k++){
45                     int dis = abs(p[i - 1][k].pos - p[i][j].pos);
46                     f[i][j] = min(f[i][j], f[i - 1][k] + dis + p[i][j].v);
47                 }
48             }
49         }
50         int ans = INF;
51         for (int i = 1; i <= t; i++){
52             int dis = abs(p[c][i].pos - L);
53             ans = min(ans, f[c][i] + dis);
54         }
55         printf("%d\n", ans);
56     }
57 }

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