poj3140Contestants Division

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这叫树形DP吗。。？断开某条边 求剩下两颗树数权值和的差最小

dfs一遍 枚举边 查了n久 wa n次  dp数组没初始化。。

在poj上1A感觉应该挺爽

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 using namespace std;

 7 #define N 100010

 8 #define M 1000010

 9 #define LL long long

10 LL dp[M<<1],p[N];

11 struct node

12 {

13     int u,v,next;

14 }ed[M<<1];

15 int head[N],t,n,m;

16 void init()

17 {

18     t = 1;

19     memset(head,-1,sizeof(head));

20 }

21 void add(int u,int v)

22 {

23     ed[t].u = u;

24     ed[t].v = v;

25     ed[t].next = head[u];

26     head[u] = t++;

27 }

28 LL dfs(int pre,int u,int e)

29 {

30     int i;

31     LL sum=p[u];

32     for(i = head[u] ; i != -1 ; i = ed[i].next)

33     {

34         int v = ed[i].v;

35         if(v==pre)

36         continue;

37         sum+=dfs(u,v,i);

38     }

39     dp[e] = sum;

40     return dp[e];

41 }

42 int main()

43 {

44     int i,kk=0;

45     while(scanf("%d%d",&n,&m)!=EOF)

46     {

47         if(n==0&&m==0)

48         break;

49         init();kk++;

50         memset(dp,0,sizeof(dp));

51         LL s=0;

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

53         {

54             scanf("%lld",&p[i]);

55             s+=p[i];

56         }

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

58         {

59             int u,v;

60             scanf("%d%d",&u,&v);

61             add(u,v);

62             add(v,u);

63         }

64         dfs(-1,i,0);

65         LL ans = s,o;

66         for(i = 0 ; i < t ; i++)

67         {

68             if(s-dp[i]>dp[i])

69             o = s-dp[i]-dp[i];

70             else

71             o = dp[i]-(s-dp[i]);

72             if(o<ans)

73             ans = o;

74         }

75         printf("Case %d: ",kk);

76         printf("%lld\n",ans);

77     }

78     return 0;

79 }

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