zoukankan      html  css  js  c++  java
  • HDU5137-最短路-删点

    How Many Maos Does the Guanxi Worth

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
    Total Submission(s): 3290    Accepted Submission(s): 1291


    Problem Description
    "Guanxi" is a very important word in Chinese. It kind of means "relationship" or "contact". Guanxi can be based on friendship, but also can be built on money. So Chinese often say "I don't have one mao (0.1 RMB) guanxi with you." or "The guanxi between them is naked money guanxi." It is said that the Chinese society is a guanxi society, so you can see guanxi plays a very important role in many things.

    Here is an example. In many cities in China, the government prohibit the middle school entrance examinations in order to relief studying burden of primary school students. Because there is no clear and strict standard of entrance, someone may make their children enter good middle schools through guanxis. Boss Liu wants to send his kid to a middle school by guanxi this year. So he find out his guanxi net. Boss Liu's guanxi net consists of N people including Boss Liu and the schoolmaster. In this net, two persons who has a guanxi between them can help each other. Because Boss Liu is a big money(In Chinese English, A "big money" means one who has a lot of money) and has little friends, his guanxi net is a naked money guanxi net -- it means that if there is a guanxi between A and B and A helps B, A must get paid. Through his guanxi net, Boss Liu may ask A to help him, then A may ask B for help, and then B may ask C for help ...... If the request finally reaches the schoolmaster, Boss Liu's kid will be accepted by the middle school. Of course, all helpers including the schoolmaster are paid by Boss Liu.

    You hate Boss Liu and you want to undermine Boss Liu's plan. All you can do is to persuade ONE person in Boss Liu's guanxi net to reject any request. This person can be any one, but can't be Boss Liu or the schoolmaster. If you can't make Boss Liu fail, you want Boss Liu to spend as much money as possible. You should figure out that after you have done your best, how much at least must Boss Liu spend to get what he wants. Please note that if you do nothing, Boss Liu will definitely succeed.
     
    Input
    There are several test cases.

    For each test case:

    The first line contains two integers N and M. N means that there are N people in Boss Liu's guanxi net. They are numbered from 1 to N. Boss Liu is No. 1 and the schoolmaster is No. N. M means that there are M guanxis in Boss Liu's guanxi net. (3 <=N <= 30, 3 <= M <= 1000)

    Then M lines follow. Each line contains three integers A, B and C, meaning that there is a guanxi between A and B, and if A asks B or B asks A for help, the helper will be paid C RMB by Boss Liu.

    The input ends with N = 0 and M = 0.

    It's guaranteed that Boss Liu's request can reach the schoolmaster if you do not try to undermine his plan.
     
    Output
    For each test case, output the minimum money Boss Liu has to spend after you have done your best. If Boss Liu will fail to send his kid to the middle school, print "Inf" instead.
     
    Sample Input
    4 5 1 2 3 1 3 7 1 4 50 2 3 4 3 4 2 3 2 1 2 30 2 3 10 0 0
          可以选择删掉图中的一个点后使得最短路尽可能的大,点数很少,枚举所有点依次删除,然后跑n-2次最短路就好了。
        
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<algorithm>
     6 #include<map>
     7 #include<set>
     8 #include<vector>
     9 #include<functional>
    10 using namespace std;
    11 #define LL long long 
    12 #define pii pair<int,int>
    13 #define mp make_pair
    14 #define inf 0x3f3f3f3f
    15 struct Edge{
    16     int v,w,next;
    17 }e[2050];
    18 int first[50],tot;
    19 void add(int u,int v,int w){
    20     e[tot].v=v;
    21     e[tot].w=w;
    22     e[tot].next=first[u];
    23     first[u]=tot++;
    24 }
    25 int vis[50],d[50],n,m;
    26 void dij(int no){
    27     memset(vis,0,sizeof(vis));
    28     memset(d,inf,sizeof(d));
    29     priority_queue<pii,vector<pii>,greater<pii> >q;
    30     d[1]=0;
    31     q.push(mp(0,1));
    32     while(!q.empty()){
    33         int u=q.top().second;
    34         q.pop();
    35         if(vis[u]) continue;
    36         vis[u]=1;
    37         for(int i=first[u];i+1;i=e[i].next){
    38             if(e[i].v==no) continue;
    39             if(d[e[i].v]>d[u]+e[i].w){
    40                 d[e[i].v]=d[u]+e[i].w;
    41                 q.push(mp(d[e[i].v],e[i].v));
    42             }
    43         }
    44     }
    45 }
    46 int main()
    47 {
    48     int i,j,k,ans;
    49     int u,v,w;
    50     while(cin>>n>>m&&(n||m)){ans=-1;
    51         memset(first,-1,sizeof(first));
    52         tot=0;
    53         while(m--){
    54             scanf("%d%d%d",&u,&v,&w);
    55             add(u,v,w);
    56             add(v,u,w);
    57         }
    58         for(i=2;i<n;++i){
    59             dij(i);
    60             ans=max(ans,d[n]);
    61         }
    62         if(ans==inf) puts("Inf");
    63         else cout<<ans<<endl;
    64     }
    65     return 0;
    66 }
    Sample Output
    50 Inf
     
  • 相关阅读:
    微软面试问题 情商测试
    SQL游标使用实例
    如何减小与“大牛”的差距
    Dotnet面试题
    排序算法对冒泡排序的优化改进算法
    一个SQL实现薪水大于所在部门平均薪水的员工
    ASP.NET中TextBox设置为Readonly后无法取值的解决办法
    jQuery.Autocomplete实现自动完成功能(详解)
    php发送get、post请求的几种方法
    ISO Latin1字符集
  • 原文地址:https://www.cnblogs.com/zzqc/p/8824683.html
Copyright © 2011-2022 走看看