zoukankan      html  css  js  c++  java
  • find the longest of the shortest (hdu 1595 SPFA+枚举)

    find the longest of the shortest

    Time Limit: 1000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1916    Accepted Submission(s): 668


    Problem Description
    Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another.
    Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed.
    Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.
     

    Input
    Each case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N.
    In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.
     

    Output
    In the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.
     

    Sample Input
    5 6 1 2 4 1 3 3 2 3 1 2 4 4 2 5 7 4 5 1 6 7 1 2 1 2 3 4 3 4 4 4 6 4 1 5 5 2 5 2 5 6 5 5 7 1 2 8 1 4 10 2 3 9 2 4 10 2 5 1 3 4 7 3 5 10
     

    Sample Output
    11 13 27
     
    先求整个的最短路,并记录下最短路经,再分别枚举删除最短路经上的每一条边。并再次求最短路,答案即最长的那条最短路。


    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <string>
    #include <map>
    #include <stack>
    #include <vector>
    #include <set>
    #include <queue>
    #pragma comment (linker,"/STACK:102400000,102400000")
    #define maxn 1010
    #define MAXN 2005
    #define mod 1000000009
    #define INF 0x3f3f3f3f
    #define pi acos(-1.0)
    #define eps 1e-6
    typedef long long ll;
    using namespace std;
    
    struct Edge
    {
        int v,w;
        int next;
    }edge[maxn*maxn];
    
    int head[maxn],inq[maxn],inqtime[maxn],dist[maxn];
    int path[maxn];
    int N,num,m,a[maxn],ok;
    bool flag[maxn];
    
    void addedge(int u,int v,int w)
    {
        edge[num].v=v;
        edge[num].w=w;
        edge[num].next=head[u];
        head[u]=num++;
        edge[num].v=u;
        edge[num].w=w;
        edge[num].next=head[v];
        head[v]=num++;
    }
    
    int SPFA(int s,int n)
    {
        int i;
        memset(inq,0,sizeof(inq));
        memset(dist,INF,sizeof(dist));
        if (ok)
        {
            for (int i=0;i<=n;i++)
                path[i]=-1;
        }
        path[s]=0;
        queue<int>Q;
        inq[s]=1;
        dist[s]=0;
        Q.push(s);
        while (!Q.empty())
        {
            int u=Q.front();
            Q.pop();
            inq[u]=0;
            for (i=head[u];i;i=edge[i].next)
            {
                if (dist[ edge[i].v ]>dist[u]+edge[i].w)
                {
                    if (ok)
                        path[ edge[i].v ]=u;
                    dist[ edge[i].v ]=dist[u]+edge[i].w;
                    if (!inq[ edge[i].v ])
                    {
                        inq[ edge[i].v ]=1;
                        Q.push(edge[i].v);
                    }
                }
            }
        }
        return dist[N];
    }
    
    int main()
    {
        int i,j;
        while (~scanf("%d%d",&N,&m))
        {
            int u,v,w;
            memset(head,0,sizeof(head));
            num=1;
            ok=1;
            for (i=0;i<m;i++)
            {
                scanf("%d%d%d",&u,&v,&w);
                addedge(u,v,w);
            }
    //        printf("num=%d
    ",num);
            SPFA(1,N);
            ok=0;
            j=N;
            int ans=-1;
            while (path[j]>0)
            {
                int x,y,value;
                for (int i=head[j];i!=0;i=edge[i].next)
                {
                    if (edge[i].v==path[j])
                    {
                        x=i;
                        value=edge[i].w;
                        edge[i].w=INF;
                        break;
                    }
                }
                for (int i=head[path[j]];i!=0;i=edge[i].next)
                {
                    if (edge[i].v==j)
                    {
                        y=i;
                        edge[i].w=INF;
                        break;
                    }
                }
                ans=max(ans,SPFA(1,N));
                edge[x].w=value;
                edge[y].w=value;
    //            printf("j=%d
    ",j);
                j=path[j];
            }
    //        printf("
    %d
    ",path[1]);
            printf("%d
    ",ans);
        }
        return 0;
    }
    


  • 相关阅读:
    ASP.NET MVC 3:缓存功能的设计问题
    Java开发和运行环境的搭建
    有关TSQL的10个好习惯
    j2me开发J2ME常用的手机厂商SDK和下载地址
    j2me开发Eclipse开发J2ME程序之Hello World
    winform需要缓存吗?webform的缓存的作用?
    学习J2ME编程需要掌握的七种技术
    下一代工作流大曝光:将支持C#表达式
    j2me 出色的开源项目
    eclipse和Myeclipse什么关系
  • 原文地址:https://www.cnblogs.com/yangykaifa/p/7206359.html
Copyright © 2011-2022 走看看