POJ 2378 Tree Cutting 3140 Contestants Division (简单树形dp)

POJ 2378 Tree Cutting:题意求删除哪些单点后产生的森林中的每一棵树的大小都小于等于原树大小的一半

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 const int maxn=10010;
 8 const int maxx=1<<30;
 9 int t,n;
10 int f[maxn]={};
11 int vis[maxn]={};
12 struct nod{
13     int y;
14     int next;
15 }e[2*maxn]={};
16 int head[maxn]={},tot=0;
17 int ans[maxn]={},size=0,sum=0;
18 void init(int x,int y){
19     e[++tot].y=y;
20     e[tot].next=head[x];
21     head[x]=tot;
22 }
23 void dfs(int x){
24     int y;
25     vis[x]=1;
26     int tmp=0;
27     for(int i=head[x];i;i=e[i].next){
28         y=e[i].y;
29         if(!vis[y]){
30             dfs(y);
31             f[x]+=f[y]+1;
32             tmp=max(f[y]+1,tmp);
33         }
34     }
35     tmp=max(tmp,n-f[x]-1);
36     if(tmp<size){
37         ans[++sum]=x;
38     }
39 }
40 int main(){
41     scanf("%d",&n);
42     size=n/2+1;
43     int x,y;
44     for(int i=1;i<n;i++){
45         scanf("%d%d",&x,&y);
46         init(x,y);
47         init(y,x);
48     }
49     dfs(1);
50     sort(ans+1,ans+1+sum);
51     for(int i=1;i<=sum;i++){
52         printf("%d
",ans[i]);
53     }
54     return 0;
55 }

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POJ 3140 Contestants Division:题意删除某边后剩余两个联通块节点数的差最小,输出这个差,

要用longlong,记得把上限设大...

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 const int maxn=100010;
 8 long long maxx=2000000000000000000LL;
 9 int t,n,m;
10 long long f[maxn]={},v[maxn]={};
11 int vis[maxn]={};
12 struct nod{
13     int y;
14     int next;
15 }e[2*maxn]={};
16 int head[maxn]={},tot;
17 long long ans,sum,bign;
18 void init(int x,int y){
19     e[++tot].y=y;
20     e[tot].next=head[x];
21     head[x]=tot;
22 }
23 inline long long mabs(long long x){
24     if(x<0){
25         return -x;
26     }
27     return x;
28 }
29 void dfs(int x){
30     int y;
31     long long tmp=0;
32     vis[x]=1;
33     for(int i=head[x];i;i=e[i].next){
34         y=e[i].y;
35         if(!vis[y]){
36             dfs(y);
37             f[x]+=f[y];
38             tmp=mabs(bign-f[y]-f[y]);
39             ans=min(ans,tmp);
40         }
41     }
42     f[x]+=v[x];
43 }
44 void yu(){
45     bign=sum=0;
46     tot=0;
47     ans=maxx;
48     memset(f,0,sizeof(f));
49     memset(vis,0,sizeof(vis));
50     memset(head,0,sizeof(head));
51     memset(v,0,sizeof(v));
52 }
53 int main(){
54     for(int zz=1;;zz++){
55         yu();
56         scanf("%d%d",&n,&m);
57         if(n==0&&m==0){
58             break;
59         }
60         int x,y;
61         for(int i=1;i<=n;i++){
62             scanf("%lld",&v[i]);
63             bign+=v[i];
64         }
65         for(int i=1;i<=m;i++){
66             scanf("%d%d",&x,&y);
67             init(x,y);
68             init(y,x);
69         }
70         dfs(1);
71         printf("Case %d: %lld
",zz,ans);
72     }
73     return 0;
74 }

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