Strategic Game
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7477 Accepted Submission(s): 3558
Problem 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.
The input file contains several data sets in text format. Each data set represents a tree with the following description:
Your program should find the minimum number of soldiers that Bob has to put for a given tree.
The input file 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
or
node_identifier:(0)
The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.
For example for the tree:
the solution is one soldier ( at the node 1).
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 table:
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
题意：
给你一棵树，每个节点，可以放一个人，这个人的视野范围是他到父节点和子节点的路，求用最少的人是他们的视野范围为整棵树。
思路：
就是一个点选与不选的问题；dp[root][1] 表示选择root点时最少用多少， dp[root][0]表示不选择root点时最少用多 少
状态转移方程为：
dp[root][0] += dp[son][1];
dp[root][1] += min(dp[son][0], dp[son][1]);
dp[root][1] += min(dp[son][0], dp[son][1]);
AC代码：
1 # include <bits/stdc++.h> 2 using namespace std; 3 const int MAX = 1550; 4 struct Node 5 { 6 int to; 7 int next; 8 }tree[MAX * 2]; 9 int head[MAX]; 10 int tol = 0; 11 int dp[MAX][2]; 12 void add(int a, int b) 13 { 14 tree[tol].to = b; 15 tree[tol].next = head[a]; 16 head[a] = tol++; 17 18 tree[tol].to = a; 19 tree[tol].next = head[b]; 20 head[b] = tol++; 21 } 22 void dfs(int root, int f) 23 { 24 for(int i = head[root]; i != -1; i = tree[i].next) 25 { 26 int son = tree[i].to; 27 if(son == f) 28 continue; 29 dfs(son, root); 30 dp[root][0] += dp[son][1]; 31 dp[root][1] += min(dp[son][0], dp[son][1]); 32 } 33 } 34 int main() 35 { 36 int n; 37 while(scanf("%d", &n) != EOF) 38 { 39 memset(head, -1, sizeof(head)); 40 memset(dp, 0, sizeof(dp)); 41 tol = 0; 42 for(int i = 1; i <= n; i++) 43 { 44 int a, len; 45 scanf("%d:(%d)", &a, &len); 46 for(int j = 0; j < len; j++) 47 { 48 int b; 49 scanf("%d", &b); 50 add(a + 1, b + 1); 51 } 52 dp[i][1] = 1; 53 } 54 int root = 1; 55 dfs(root, -1); 56 printf("%d ", min(dp[root][1], dp[root][0])); 57 } 58 return 0; 59 }