Strategic game
| Time Limit: 2000MS | Memory Limit: 10000K | |
| Total Submissions: 6260 | Accepted: 2883 |
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?
the solution is one soldier ( at the node 1).
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_identifiernumber_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
12
题意：
在树上的节点上放尽量少的士兵，监视所有的节点。。
分析：类似01背包，该节点只有 放或不放 士兵的两种选择！
状态：
dp[i][0] 表示以i为根节点，i不放士兵最优值。
dp[i][1] 表示以i为根节点，i放士兵的最优值。
状态转移方程
dp[i][0]=Σ dp[j][1] j为i子节点
dp[i][1]=Σ min(dp[j][1],dp[j][0])
边界：
全部节点初始化
dp[i][0] = 0;
dp[i][1]= 1;
#include <stdio.h> #include <vector> using namespace std; #define INF 99999999 vector <int> edge[2005]; int dp[2005][2]; int n; void DFS(int t,int p) { int i,k; if (edge[t].size()==1 && p!=-1) { dp[t][0]=0; dp[t][1]=1; return; } dp[t][0]=dp[t][1]=0; for (i=0;i<edge[t].size();i++) { k=edge[t][i]; if (k==p) continue; DFS(k,t); dp[t][0]+=dp[k][1]; dp[t][1]+=min(dp[k][0],dp[k][1]); } dp[t][1]++; } int main() { int i,j,x,y,k; while(scanf("%d",&n)!=EOF) { for (i=0;i<n;i++) { edge[i].clear(); } for (i=0;i<n;i++) { scanf("%d:(%d)",&x,&k); while(k--) { scanf("%d",&y); edge[x].push_back(y); edge[y].push_back(x); } } DFS(0,-1); printf("%d\n",min(dp[0][0],dp[0][1])); } return 0; }