URAL1389. Roadworks（dp)

1389

算个简单的树形DP吧 不知道是不是数据太水 竟然一A了 

就是对于当前节点有没有被选中就行选最优 有没有被选中的意思是有没有与它相连的边被选中 

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 using namespace std;

 9 #define N 100010

10 vector<int>d[N];

11 int dp[N][3],g;

12 struct node

13 {

14     int u,v;

15 }p[N];

16 int dfs(int u,int v,int f)

17 {

18     if(dp[v][f]!=-1)

19     return dp[v][f];

20     int i;

21     int s[2][2] = {0};

22     int flag =0,x=-1;

23     for(i = 0 ; i < d[v].size() ; i++)

24     {

25         int tv = d[v][i];

26         if(tv==u) continue;

27         flag = 1;

28         int s1 = dfs(v,tv,0);

29         if(f)

30         {

31             s[(i+1)%2][0] = s[i%2][0]+s1;

32         }

33         else

34         {

35             int s2 = dfs(v,tv,1)+1;

36             s[(i+1)%2][0] = s[i%2][0]+s1;

37             if(s[i%2][1]+s1<s[i%2][0]+s2)

38             {

39                 s[(i+1)%2][1] = s[i%2][0]+s2;

40                 x = tv;

41             }

42             else

43             s[(i+1)%2][1] = s[i%2][1]+s1;

44         }

45     }

46     if(!flag)

47     {

48         return dp[v][f] = 0;

49     }

50     if(x!=-1)

51     {

52         g++;

53         p[g].u = v;

54         p[g].v = x;

55     }

56     int maxz = 0;

57     for(i = 0 ; i < 2 ; i++)

58     maxz = max(maxz,max(s[0][i],s[1][i]));

59     return dp[v][f] = maxz;

60 }

61 int main()

62 {

63     int i,n,m;

64     while(cin>>n>>m)

65     {

66         memset(dp,-1,sizeof(dp));

67         for(i = 0 ; i <= n ;i++)

68         d[i].clear();

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

70         {

71             int u,v;

72             cin>>u>>v;

73             d[u].push_back(v);

74             d[v].push_back(u);

75         }

76         dfs(-1,1,0);

77         cout<<dp[1][0]<<endl;

78         for(i = 1; i <= g ; i++)

79         cout<<p[i].u<<" "<<p[i].v<<endl;

80     }

81     return 0;

82 }

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