POJ1463Strategic game题解动态规划DP

Strategic game

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 2096		Accepted: 870

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

入门树状DP
题目描述了一个树形结构，在结点上可安放0个或1个士兵，
要求在满足每条边的两端至少有一个士兵的条件下求得士兵最少数

深度优先搜索遍历树的每一个节点，对于一个节点u，存在两种状态，
以1表示放置，0表示不放置，
d[u][1]表示节点u放置士兵情况下以u为节点子树的最优值，
同理d[u][0]表示不放置为最优值，
转移方程为：
for(v 属于 u的子节点) d[u][1] += max(d[v][0], d[v][1])
for(v 属于 u的子节点) d[u][0] += d[v][1]

最后结果为根节点放置和不放置情况中的最优值

#include<cstdio> #include<cstring> #include<vector> #include<algorithm> using namespace std; int dp[1505][2],n; vector<int> v[1505]; void dfs(int p) { int i,len=v[p].size(); for(i=0;i<len;i++) dfs(v[p][i]); if(!len)//叶子结点 { dp[p][0]=0; dp[p][1]=1; } else { for(i=0;i<len;i++) { int j=v[p][i]; dp[p][1]+=min(dp[j][1],dp[j][0]); dp[p][0]+=dp[j][1]; } dp[p][1]++; } } int main() { while(scanf("%d",&n)==1) { memset(dp,0,sizeof(dp)); int i,j,k,p,m,x,f[1505]={0}; for(i=0;i<n;i++) v[i].clear(); for(i=0;i<n;i++) { scanf("%d:(%d)",&p,&m); while(m--) { scanf("%d",&x); v[p].push_back(x); f[x]=1; } } for(i=0;i<n;i++) if(!f[i])//树根 break; dfs(i); printf("%d/n",min(dp[i][0],dp[i][1])); } }