zoukankan      html  css  js  c++  java
  • (状压dp)ABC 067 F : Mole and Abandoned Mine

    Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1through N and M edges. The i-th edge connects Vertices ai and bi, and it costs ciyen (the currency of Japan) to remove it.

    Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

    Constraints

     

    • 2≤N≤15
    • N−1≤MN(N−1)⁄2
    • 1≤ai,biN
    • 1≤ci≤106
    • There are neither multiple edges nor self-loops in the given graph.
    • The given graph is connected.

    Input

     

    Input is given from Standard Input in the following format:

    N M
    a1 b1 c1
    :
    aM bM cM
    

    Output

     

    Print the answer.

    Sample Input 1

     

    4 6
    1 2 100
    3 1 100
    2 4 100
    4 3 100
    1 4 100
    3 2 100
    

    Sample Output 1

     

    200
    

    By removing the two edges represented by the red dotted lines in the figure below, the objective can be achieved for a cost of 200 yen.

    45c15676bb602ca3b762561fc014ecd0.png

    Sample Input 2

     

    2 1
    1 2 1
    

    Sample Output 2

     

    0
    

    It is possible that there is already only one path from Vertex 1 to Vertex N in the beginning.

    Sample Input 3

     

    15 22
    8 13 33418
    14 15 55849
    7 10 15207
    4 6 64328
    6 9 86902
    15 7 46978
    8 14 53526
    1 2 8720
    14 12 37748
    8 3 61543
    6 5 32425
    4 11 20932
    3 12 55123
    8 2 45333
    9 12 77796
    3 9 71922
    12 15 70793
    2 4 25485
    11 6 1436
    2 7 81563
    7 11 97843
    3 1 40491
    

    Sample Output 3

     

    133677

    同时进行两种递推:1、加入单独1点 2、加入一与当前无交集的点集    
    get: line50 如何取得补集的所有子集。
     1 #include <bits/stdc++.h>
     2 typedef long long ll;
     3 using namespace std;
     4 const int MAX=1e5+5;
     5 const int INF=1e9;
     6 int n,m;
     7 int edge[20][20];
     8 int total;
     9 int cost[1<<16][16],dp[1<<16][16];//dp[s][x]记录达到某点集s,最后“末端”(继续连接其余点的唯一点)为x时余下权值和最大的边的情况
    10 int inner[1<<16];//内部的边
    11 int main()
    12 {
    13     scanf("%d%d",&n,&m);
    14     for(int i=1;i<=m;i++)//读入数据
    15     {
    16         int x,y,price;
    17         scanf("%d%d%d",&x,&y,&price);
    18         --x;--y;
    19         edge[x][y]=edge[y][x]=price;
    20     }
    21     total=1<<n;
    22     for(int i=0;i<total;i++)//计算某点集内部所有边权值之和
    23         for(int a=0;a<n;a++)
    24             if(i&(1<<a))
    25                 for(int b=0;b<a;b++)
    26                     if(i&(1<<b))
    27                         if(edge[a][b])
    28                             inner[i]+=edge[a][b];
    29     for(int i=0;i<total;i++)//计算某点集到点集外某点所有边的权值之和
    30         for(int a=0;a<n;a++)
    31             if(!(i&(1<<a)))
    32                 for(int b=0;b<n;b++)
    33                     if((i&(1<<b))&&edge[a][b])
    34                         cost[i][a]+=edge[a][b];
    35     memset(dp,-1,sizeof(dp));
    36     dp[1][0]=0;//只有1个点的集合dp值显然为0
    37     for(int i=0;i<total;i++)
    38     {
    39 
    40         for(int a=0;a<n;a++)
    41         {
    42             if((i&(1<<a))&&dp[i][a]!=-1)
    43             {
    44                 for(int b=0;b<n;b++)
    45                     if(!(i&(1<<b)))/*只加入一个点*/
    46                         if(edge[a][b])
    47                             dp[i|(1<<b)][b]=max(dp[i|(1<<b)][b],dp[i][a]+edge[a][b]);
    48                 /*加入一个点集*/
    49                 int left=total-1-i;
    50                 for(int now=left;now!=0;now=(now-1)&left)//用此循环得到所有
    51                 {
    52                     int num=inner[now]+cost[now][a];
    53                     dp[i|now][a]=max(dp[i|now][a],dp[i][a]+num);
    54                 }
    55             }
    56         }
    57     }
    58     printf("%d
    ",inner[total-1]-dp[total-1][n-1]);
    59     return 0;
    60 }
  • 相关阅读:
    Android 应用开发耗电量控制。。
    android优化从网络中加载图片速度。。
    SpringMVC 配置多视图解析器(velocity,jsp)
    linux mysql定时备份并压缩
    linux mysql定时备份并压缩
    mysql选择上一条、下一条数据记录,排序上移、下移、置顶
    MIT-CBCL Car Database 车辆训练数据集
    两个对象值相同(x.equals(y) == true),但却可有不同的hash code,这句话对不对?
    js实现ArrayList功能
    JXL.jar简单封装Excel读写操作
  • 原文地址:https://www.cnblogs.com/quintessence/p/7197078.html
Copyright © 2011-2022 走看看