zoukankan      html  css  js  c++  java
  • 图论(二分图,KM算法):HDU 3488 Tour

    Tour

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
    Total Submission(s): 2628    Accepted Submission(s): 1285


    Problem Description
    In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)
    Every city should be just in one route.
    A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
    The total distance the N roads you have chosen should be minimized.
     
    Input
    An integer T in the first line indicates the number of the test cases.
    In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
    It is guaranteed that at least one valid arrangement of the tour is existed.
    A blank line is followed after each test case.
     

    Output

    For each test case, output a line with exactly one integer, which is the minimum total distance.
     
    Sample Input
    1
    6 9
    1 2 5
    2 3 5
    3 1 10
    3 4 12
    4 1 8
    4 6 11
    5 4 7
    5 6 9
    6 5 4
     
    Sample Output
    42
     
       这题就是KM算法的模板。
     1 #include <iostream>
     2 #include <cstring>
     3 #include <cstdio>
     4 using namespace std;
     5 const int maxn=220;
     6 const int INF=2147483647;
     7 int w[maxn][maxn],sx[maxn],sy[maxn],lx[maxn],ly[maxn];
     8 int match[maxn],slack[maxn];
     9 int n,m;
    10 
    11 bool Search(int x){
    12     sx[x]=true;
    13     for(int y=1;y<=n;y++){
    14         if(sy[y])continue;//?
    15         int t=lx[x]+ly[y]-w[x][y];
    16         if(t)
    17             slack[y]=min(slack[y],t);
    18         else{
    19             sy[y]=true;
    20             if(!match[y]||Search(match[y])){
    21                 match[y]=x;
    22                 return true;
    23             }
    24         }
    25     }
    26     return false;
    27 }
    28 
    29 int KM(){
    30     memset(lx,0x80,sizeof(lx));
    31     memset(ly,0,sizeof(ly));
    32     for(int i=1;i<=n;i++)
    33         for(int j=1;j<=n;j++)
    34             lx[i]=max(lx[i],w[i][j]);
    35     
    36     for(int i=1;i<=n;i++){
    37         memset(slack,127,sizeof(slack));
    38         while(true){
    39             memset(sx,0,sizeof(sx));
    40             memset(sy,0,sizeof(sy));
    41             if(Search(i))
    42                 break;
    43             int minn=INF;
    44             for(int j=1;j<=n;j++)
    45                 if(!sy[j]&&minn>slack[j])
    46                     minn=slack[j];
    47             
    48             for(int j=1;j<=n;j++)
    49                 if(sx[j])
    50                     lx[j]-=minn;
    51                     
    52             for(int j=1;j<=n;j++)
    53                 if(sy[j])
    54                     ly[j]+=minn;
    55                 else 
    56                     slack[j]-=minn;
    57         }    
    58     }    
    59     int ret=0;
    60     for(int i=1;i<=n;i++)
    61         ret+=w[match[i]][i];
    62     return ret;    
    63 }
    64 int main(){
    65     int T;
    66     scanf("%d",&T);
    67     while(T--){
    68         scanf("%d%d",&n,&m);
    69         memset(w,0x80,sizeof(w));
    70         memset(match,0,sizeof(match));
    71         for(int i=1,a,b,c;i<=m;i++){
    72             scanf("%d%d%d",&a,&b,&c);
    73             w[a][b]=max(w[a][b],-c);
    74         }
    75         printf("%d
    ",-KM());
    76     }
    77     return 0;
    78 }
    尽最大的努力,做最好的自己!
  • 相关阅读:
    MapiRule例子
    P/invoke in .NET Compact Framework
    MFC C++类型学习
    Windows Mobile上实现可拖动的窗口
    在Wince下使用钩子函数
    VC++动态链接库编程之MFC规则DLL
    VC++动态链接库编程之DLL典型实例
    Using keyboard hooks in WinCE
    Override VK_TTALK & VK_TEND
    Getphonenumber获得电话号码的例子
  • 原文地址:https://www.cnblogs.com/TenderRun/p/5330061.html
Copyright © 2011-2022 走看看