ZOJ Monthly, June 2012 - J Escape Time II - zoj 3620

题目说要在规定时间内，拿走最大价值。因为房间n <= 10  太小了，所以直接枚举 走过哪几个房间，然后dp 递推下去就可以了

dp[i][j] 走过i状态的房间，现在在第j个房间上面花费的最小时间。

View Code

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<algorithm>

 4 #include<cstring>

 5 #include<cmath>

 6 using namespace std;

 7 #define N 12

 8 #define M 110

 9 #define inf 1000000000

10 typedef long long LL;

11 int v[N], val[1<<N];

12 struct node {

13     int v, next, l;

14 }e[M];

15 int mpp[N][N], dp[1<<N][N];

16 int main() {

17     int n, m, t, s, e;

18     while (scanf("%d%d%d", &n, &m, &t) != EOF) {

19         scanf("%d%d", &s, &e);

20         for (int i = 0; i < n; ++i) {

21             scanf("%d", &v[i]);

22             for (int j = 0; j < n; ++j)

23                 mpp[i][j] = inf;

24             mpp[i][i] = 0;

25         }

26 

27         for (int i = 0; i < 1<<n; ++i) {

28             for (int j = 0; j < n; ++j)

29                 dp[i][j] = inf;

30             val[i] = 0;

31             int temp = i, k = 0;

32             while (temp) {

33                 if (temp&1) val[i] += v[k];

34                 k++;

35                 temp >>= 1;

36             }

37         }

38 

39         while (m--) {

40             int a, b, c;

41             scanf("%d%d%d", &a, &b, &c);

42             mpp[a][b] = mpp[b][a] = min(mpp[a][b], c);

43         }

44         for (int k = 0; k < n; ++k) for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j)

45             mpp[i][j] = min(mpp[i][j], mpp[i][k]+mpp[k][j]);

46         dp[1<<s][s] = 0;

47         for (int i = 1; i < 1<<n; ++i) {

48             for (int j = 0; j < n; ++j) {

49                 if (dp[i][j] > t) continue;

50                 for (int k = 0; k < n; ++k) {

51                     int state = i|1<<k;

52                     dp[state][k] = min(dp[state][k], mpp[j][k]+dp[i][j]);

53                 }

54             }

55         }

56         int ans = 0;

57         for (int i = 1; i < 1<<n; ++i)

58             if (dp[i][e] <= t)

59                 ans = max(val[i], ans);

60         printf("%d\n", ans);

61     }

62     return 0;

63 }