USACO 4.4 Pollutant Control

第一二问，可以通过*1001+1，最后ans/1001 ans%1001来解决，第三问，看题解后完全无想法，然后找了一种水过的办法，水过了。。这种做法应该是错的。

 1 /*

 2  ID:cuizhe

 3  LANG: C++

 4  TASK: milk6

 5  */

 6 #include <cstdio>

 7 #include <cstring>

 8 #include <cmath>

 9 #include <queue>

10 using namespace std;

11 #define LL long long

12 LL INF;

13 LL flow[201][201];

14 LL low[201];

15 int path[201],used[201];

16 int x[1001],y[1001];

17 int str,end,n,m;

18 LL bfs()

19 {

20     int t,i;

21     memset(path,-1,sizeof(path));

22     memset(used,0,sizeof(used));

23     used[str] = 1;

24     queue<int> que;

25     que.push(str);

26     low[str] = INF;

27     while(!que.empty())

28     {

29         t = que.front();

30         que.pop();

31         if(t == end) break;

32         for(i = 1; i <= n; i ++)

33         {

34             if(i != str&&!used[i]&&flow[t][i])

35             {

36                 low[i] = low[t] < flow[t][i] ? low[t]:flow[t][i];

37                 used[i] = 1;

38                 path[i] = t;

39                 que.push(i);

40             }

41         }

42     }

43     if(path[end] == -1)

44         return -1;

45     else

46         return low[end];

47 }

48 LL EK()

49 {

50     LL res,now,ans = 0;

51     while((res = bfs()) != -1)

52     {

53         ans += res;

54         now = end;

55         while(now != str)

56         {

57             flow[now][path[now]] += res;

58             flow[path[now]][now] -= res;

59             now = path[now];

60         }

61     }

62     return ans;

63 }

64 int main()

65 {

66     int i,sv,ev;

67     LL ans,w;

68     INF = (LL)9999999*9999999;

69     freopen("milk6.in","r",stdin);

70     freopen("milk6.out","w",stdout);

71     scanf("%d%d",&n,&m);

72     memset(flow,0,sizeof(flow));

73 

74     for(i = 1; i <= m; i ++)

75     {

76         scanf("%d%d%lld",&sv,&ev,&w);

77         x[i] = sv;

78         y[i] = ev;

79         flow[sv][ev] += ((LL)(w*1001+1)*500000 + i-1);

80     }

81     str = 1,end = n;

82     ans = EK();

83     printf("%lld %lld\n",ans/500500000,(ans/500000)%1001);

84     for(i = 1;i <= m;i ++)

85     {

86         if(used[x[i]]&&!used[y[i]])

87         printf("%d\n",i);

88     }

89     return 0;

90 }

View Code