Proxy（2016山东省省赛C）（最短路）（spfa)

问题 C: Proxy

时间限制: 2 Sec 内存限制: 128 MB
提交: 17 解决: 5
[提交][状态][讨论版]

题目描述

Because
of the GFW (Great Firewall), we cannot directly visit many websites,
such as Facebook, Twitter, YouTube, etc. But with the help of proxy and
proxy server, we can easily get to these website.
You have a list of several proxy servers,
some of them can be connected directly but others can’t. But you can
visit proxy servers through other proxy server by a one-way connection.
As we all know, the lag of internet visit
will decide our feelings of the visit. You have a very smart proxy
software which will find the least lag way to reach the website once you
choose a directly reachable proxy server.
You know the lag of every connection. The
lag of your visit is the all the lags in your whole connection. You want
to minimize the lag of visit, which proxy server you will choose?

输入

Multiple test cases, the first line is an integer T (T <= 100), indicating the number of test cases.
The first line of each test case is two
integers N (0 <= N <= 1000), M (0 <= M <= 20000). N is the
number of proxy servers (labeled from 1 to N). 0 is the label of your
computer and (N+1) is the label of the server of target website.
Then M lines follows, each line contains
three integers u, v, w (0 <= u, v <= N + 1, 1 <= w <= 1000),
means u can directly connect to v and the lag is w.

输出

An
integer in one line for each test case, which proxy server you will
choose to connect directly. You can only choose the proxy server which
can be connected directly from your computer.
If there are multiple choices, you should
output the proxy server with the least label. If you can’t visit the
target website by any means, output “-1” (without quotes). If you can
directly visit the website and the lag is the least, output “0” (without
quotes).

样例输入

样例输出

3

-1

0

1
【分析】要是只求最短距离那还简单，Spfa就挺好，但是要求输出你走的第一个点就有点麻烦，我们可以用一个pre数组，pre[i]存的是当走到i节点时，从0到i的
最短路中与0相连的那个节点，这样在找最短路时就可以灵活的更新了。

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define mod 1000000007

#define inf 0x3f3f3f3f

#define pi acos(-1.0)

using namespace std;

typedef long long ll;

#define RANGE 1005

#define MAX 0x3f3f3f3f

int cost[RANGE][RANGE];

int d[RANGE];

bool used[RANGE];

int n,m;

int pre[];

void spfa(int s)

{

    int i,j,now;

    for(i=;i<=n+;i++)

    {

        d[i]=MAX;

        used[i]=false;

    }

    int flag=;

    for(i=;i<=n+;i++)

    {

        if(cost[][i]<MAX)pre[i]=i;//初始化pre数组

    }

    used[s]=true;

    d[s]=;

    queue<int> q;

    q.push(s);

    while(!q.empty())

    {

        now=q.front();

        q.pop();

        for(i=;i<=n+;i++)

        {

            if(d[i]>d[now]+cost[now][i])

            {

                d[i]=d[now]+cost[now][i];

                if(used[i]==)

                {

                    if(now!=)pre[i]=pre[now];

                    used[i]==true;

                    q.push(i);

                }

            }

            if(d[i]==d[now]+cost[now][i])

            {

                    if(now!=)pre[i]=min(pre[i],pre[now]);

            }

        }

    }

}

int main()

{

    int i,j,a,b,c;

    int we;

    scanf("%d",&we);

    while(we--)

    {scanf("%d%d",&n,&m);

        if(!n&&!m) break;

        for(i=;i<=n+;++i)

        {

            for(j=;j<=i;j++)

            {

                if(i==j)  cost[i][j]=;

                else  cost[i][j]=cost[j][i]=MAX;

            }

        }

        for(i=;i<m;i++)

        {

            scanf("%d%d%d",&a,&b,&c);

            cost[a][b]=c;

        }

        spfa();

        if(d[n+]<MAX)

        {

            if(pre[n+]==n+)printf("0\n");

            else printf("%d\n",pre[n+]);

        }

        else printf("-1\n");

    }

    return ;

}