#POJ1463#Strategic Game（贪心 or Hungary or 树形DP）

Strategic game

Time Limit: 2000MS Memory Limit: 10000KTotal Submissions: 8409 Accepted: 3931

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

40:(1) 11:(2) 2 32:(0)3:(0)53:(3) 1 4 21:(1) 02:(0)0:(0)4:(0)

Sample Output

12

题意：

鲍勃喜欢玩战略游戏。现在有n个城市，他们构成了一棵树。

鲍勃可以在某些城市派一个士兵守护，该士兵可以瞭望到所有与该城市相连的边。

问鲍勃最少要派遣多少个士兵，才能把所有的边都瞭望到。 N <= 1500

这题有三种做法：贪心 匈牙利匹配 和 树形DP，现在只讲最快的贪心法

将图做成无向图。

从度为1的节点开始想，这种节点是必须看到的，但是如果放在它们的位置上，显然不如放在与之相连的位置上好

（因为这样能看见更多的士兵，而且摆放的数量也会更少，请想一想）

我们将与它们相连的点摆上士兵，然后删除度为1节点的边，并且也删除已摆放士兵的节点的边，

剩下的图中，会再次出现度为1的点，重复以上操作。

Code:

StatusAcceptedTime234msMemory1804kBLength1393LangG++Submitted2017-07-13 17:40:12Shared

#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<vector>#include<queue>#include<cstring>using namespace std;const int Max = 1500;int num[Max + 5];vector<int>Tr[Max + 5];queue<int>Q;bool vis[Max + 5];bool  getint(int & num){    char c;    int flg = 1;    num = 0;    while((c = getchar()) < '0' || c > '9')    if(c == '-')    flg = -1;    while(c >= '0' && c <= '9')    {    num = num * 10 + c - 48;    c = getchar();}    num *= flg;return 1;}int main(){int N;while(~scanf("%d", &N)){int r, x, n;for(int i = 1; i <= N; ++ i){getint(r),getint(n);for(int j =1; j <= n; ++ j)getint(x),Tr[r].push_back(x),Tr[x].push_back(r),++ num[x], ++ num[r];}if(N == 1){printf("1\n");continue;}for(int i = 0; i < N; ++ i)if(num[i] == 1)Q.push(i);int Ans = 0;while(! Q.empty()){int tmp = Q.front();Q.pop();if(vis[tmp])continue;for(int i = 0; i < (int)Tr[tmp].size(); ++ i)if(! vis[Tr[tmp][i]]){++ Ans;int u = Tr[tmp][i];vis[u] = 1;for(int j = 0; j < (int)Tr[u].size(); ++ j){-- num[Tr[u][j]];if(! vis[Tr[u][j]] && num[Tr[u][j]] == 1)Q.push(Tr[u][j]);}}}printf("%d\n", Ans);memset(num, 0, sizeof(num));memset(vis, 0, sizeof(vis));for(int i = 0; i < N; ++ i)Tr[i].clear();}return 0;}