zoukankan      html  css  js  c++  java
  • poj1463(简单树形dp)

    Strategic game
    Time Limit: 2000MS   Memory Limit: 10000K
    Total Submissions: 9313   Accepted: 4368

    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_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

    一句话题意:如果一个点上放士兵,那士兵相邻位置的路都会被看守住,现在问如果要把所有路都看守住,至少需要放多少士兵。

    设dp[x][1],表示x放士兵,dp[x][0]表示x不放士兵

    那么dp[x][1]+=min(dp[son][0],dp[son][1])

    dp[x][0]+=dp[x][1];

    。。。。。。。。。。。。。还有,记住找根,不是所以的点,都可以当根

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    
    struct my{
           int next;
           int v;
    };
    
    const int maxn=3000+10;
    int adj[maxn],fa,dp[maxn][3],du[maxn];
    my bian[maxn];
    
    void myinsert(int u,int v){
         bian[++fa].v=v;
         bian[fa].next=adj[u];
         adj[u]=fa;
    }
    
    void dfs(int x){
         if(adj[x]==-1) {
            dp[x][1]=1;
            dp[x][0]=0;
            return ;
         }
         for (int i=adj[x];i!=-1;i=bian[i].next){
            int v=bian[i].v;
            dfs(v);
            dp[x][0]+=dp[v][1];
            dp[x][1]+=min(dp[v][1],dp[v][0]);
         }
    }
    
    int main(){
        int n,u,v,m;
        while(scanf("%d",&n)!=EOF){
        memset(adj,-1,sizeof(adj));
        memset(bian,-1,sizeof(bian));
        memset(dp,0,sizeof(dp));
        memset(du,0,sizeof(du));
        fa=0;
        for (int k=1;k<=n;k++){
            scanf("%d",&u);
            dp[u][1]=1;
            scanf(":(%d)",&m);
            for (int i=1;i<=m;i++){
                scanf("%d",&v);
                myinsert(u,v);
                du[v]++;
            }
          }
          int root=0;
          for (int i=0;i<n;i++){
            if(du[i]==0) {root=i;break;}
          }
          dfs(root);
          printf("%d
    ",min(dp[root][0],dp[root][1]));
        }
    return 0;
    }
  • 相关阅读:
    codeforces cf round#505(based on vk cup 2018 final) C. Plasticine zebra
    cf round 505(div1+div2)based on VK cup 2018 final B. Weakened Common Divisor
    codeforces AIM Tec round 5(div1+div2) C. Rectangles
    codeforces cf AIM tech round5(rated for div1+div2) B. Unnatural Conditions
    C. Maximal Intersection codeforces round#506(div3)
    uva 725 Division(暴力枚举) 解题心得
    POJ 2386 Lake Counting_steven 解题心得
    UVA 1600 Patrol Robot(机器人穿越障碍最短路线BFS) 解题心得
    HDU1372:Knight Moves(BFS) 解题心得
    POJ 2255 Tree Recovery 解题心得
  • 原文地址:https://www.cnblogs.com/lmjer/p/9419177.html
Copyright © 2011-2022 走看看