HDU 1054——Strategic Game(树形DP）

Strategic Game

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4805    Accepted Submission(s): 2185

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:

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

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

 

Sample Output

12

———————————————————————分割线—————————————————————————

题目分析：

用最少的士兵将所有的道路都监控起来

这题是最小顶点覆盖，可以用二分匹配，但是今天学树形DP，走起~~

基础模板：

1、建图，用邻接表

2、dfs，遍历整个图，直到叶子，然后子问题一层一层往上合并成整个问题

额，都是废话~~~

状态转移 dp[i][0]  i节点不放士兵所需要的最少方案数,dp[i][1]  i节点放士兵的最少方案数

每个节点都有两个状态，放士兵或者不放士兵

1.如果该节点不放士兵，那么它的所有子节点必须放

2.如果该节点放士兵，那么它的子节点有两种决策，放或不放，去最小的

code^*^：

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#define maxn 1510using namespace std;struct node{    int v,next;}edge[maxn*maxn];int e;int head[maxn];int dp[maxn][2];int vis[maxn];void dfs(int u){    vis[u]=1;    dp[u][1]=1;    for(int e=head[u];e!=-1;e=edge[e].next){        int v=edge[e].v;        if(vis[v]) continue;        dfs(v);      //dfs到最底层        dp[u][1]+=min(dp[v][1],dp[v][0]);  //解决子问题        dp[u][0]+=dp[v][1];    }}void addedge(int u,int v)  //邻接表建图{    edge[e].v=v;    edge[e].next=head[u];    head[u]=e++;    edge[e].v=u;    edge[e].next=head[v];    head[v]=e++;}int main(){    int n;    while(scanf("%d",&n)!=EOF){        memset(head,-1,sizeof head);        memset(dp,0,sizeof dp);        memset(vis,0,sizeof vis);        int x,y,v,s=-1;        e=0;        while(n--){            scanf("%d:(%d)",&x,&y);            while(y--){                scanf("%d",&v);                addedge(x,v);            }            if(s==-1&&y) s=x;        }        dfs(s);        printf("%d\n",dp[s][0]>dp[s][1]?dp[s][1]:dp[s][0]); //去最小    }    return 0;}

——————vector建图

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<vector>#define maxn 1510using namespace std;vector<int> g[maxn];int dp[maxn][2];int vis[maxn];void dfs(int u){    vis[u]=1;    dp[u][1]=1;    for(int i=0;i<g[u].size();i++){        int v=g[u][i];        if(vis[v]) continue;        dfs(v);        dp[u][1]+=min(dp[v][1],dp[v][0]);        dp[u][0]+=dp[v][1];    }}int main(){    int n;    while(scanf("%d",&n)!=EOF){        memset(dp,0,sizeof dp);        memset(vis,0,sizeof vis);        int x,y,v,s=-1;        for(int i=0;i<maxn;i++){            g[i].clear();        }        while(n--){            scanf("%d:(%d)",&x,&y);            while(y--){                scanf("%d",&v);                g[x].push_back(v);                g[v].push_back(x);            }            if(s==-1&&y) s=x;        }        dfs(s);        printf("%d\n",dp[s][0]>dp[s][1]?dp[s][1]:dp[s][0]);    }    return 0;}