【POJ 1463】【hdu 1054】Strategic game（树形dp）

Strategic game Time Limit: 2000MS Memory Limit: 10000KTotal Submissions: 7983 Accepted: 3715

Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?  Your program should find the minimum number of soldiers that Bob has to put for a given tree.  For example for the tree:  the solution is one soldier ( at the node 1).

Input

The input contains several data sets in text format. Each data set represents a tree with the following description:  the number of nodes  the description of each node in the following format  node_identifier:(number_of_roads) node_identifier₁ node_identifier₂ ... node_identifier_{number_of_roads}  or  node_identifier:(0)  The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500);the number_of_roads in each line of input will no more than 10. Every edge appears only once in the input data.

Output

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following:

Sample Input

4 0:(1) 1 1:(2) 2 3 2:(0) 3:(0) 5 3:(3) 1 4 2 1:(1) 0 2:(0) 0:(0) 4:(0)

Sample Output

1 2

Source

Southeastern Europe 2000

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[题意][每个节点都可以放一个士兵，这个士兵可以瞭望到与当前节点相连的所有边，求最少需要放多少士兵使得每条边都可被瞭望] 【题解】【树形dp】 【f[i][0/1]，用f[i][0]表示当前点不放士兵，以当前点为根的子树需要多少个士兵；f[i][1]表示当前点放士兵，以当前点为根的子树需要多少个士兵。】 【方程：f[i][1]+=f[j][0](j是i的儿子)，f[i][0]=min(f[i][1],∑f[j][1]】 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,f[1510][2]; int a[3010],nxt[3010],p[1510],tot; bool b[1510]; inline void add(int x,int y) { tot++; a[tot]=y; nxt[tot]=p[x]; p[x]=tot; tot++; a[tot]=x; nxt[tot]=p[y]; p[y]=tot; } void dp(int x,int fa) { int minn=0x7fffffff,num=0; int t=0; f[x][1]=1; if(a[p[x]]==fa&&nxt[p[x]]==-1) return; for(int i=p[x];i!=-1;i=nxt[i]) if(a[i]!=fa) { dp(a[i],x); f[x][1]+=f[a[i]][0]; f[x][0]+=f[a[i]][1]; } if(f[x][1]<f[x][0]) f[x][0]=f[x][1]; return; } int main() { while(scanf("%d",&n)==1) { tot=0; memset(b,0,sizeof(b)); memset(f,0,sizeof(f)); memset(p,-1,sizeof(p)); memset(nxt,-1,sizeof(nxt)); for(int i=1;i<=n;++i) { int x,m; scanf("%d:(%d)",&x,&m); for(int j=1;j<=m;++j) { int y; scanf("%d",&y); add(x,y); b[y]=1; } } int root; for(int i=0;i<n;++i) if(!b[i]) {root=i; break; } dp(root,root); printf("%d\n",f[root][0]); } return 0; }