Strategic game
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_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
1
2
Source
题目类型：树的最小点覆盖(树形DP)
算法分析：dp[u][0]表示点u处没有设置士兵，在以u为根的子树中满足覆盖条件的最少士兵数。dp[u][1]表示点u处设置士兵，在以u为根的子树中满足覆盖条件的最少士兵数，则初始条件为：dp[i][0] = 0,dp[i][1] = 1，状态转移方程为：
dp[u][0] = Sigma(dp[v][1])
dp[u][1] += min (dp[v][0], dp[v][1])
其中v为u的子节点，最后输出min (dp[0][0], dp[0][1])即可
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1500 + 6; int dp[maxn][2], n; bool vis[maxn], edge[maxn][maxn]; inline int min (int a, int b) {return a < b ? a : b;} void dfs (int rt) { vis[rt] = true; for (int i = 0; i < n; i++) { if (!vis[i] && edge[rt][i]) { dfs (i); dp[rt][1] += min (dp[i][0], dp[i][1]); dp[rt][0] += dp[i][1]; } } } int main() { // CFF; while (scanf ("%d", &n) != EOF) { memset (edge, false, sizeof (edge)); memset (vis, false, sizeof (vis)); int u, v, nn; for (int i = 1; i <= n; i++) { scanf ("%d:(%d)", &u, &nn); for (int j = 1; j <= nn; j++) { scanf ("%d", &v); edge[u][v] = edge[v][u] = true; } } for (int i = 0; i < n; i++) dp[i][0] = 0, dp[i][1] = 1; dfs (0); printf ("%d\n", min (dp[0][0], dp[0][1])); } return 0; } |
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