XTOJ 1245 Hamiltonian Path【伪最短路，水题】

Hamiltonian Path
Accepted : 41		Submit : 96
Time Limit : 2000 MS		Memory Limit : 65536 KB

Hamiltonian Path In ICPCCamp, there are n cities and m directed roads between cities. The i-th road going from the ai-th city to the bi-th city is ci kilometers long. For each pair of cities (u,v), there can be more than one roads from u to v. Bobo wants to make big news by solving the famous Hamiltonian Path problem. That is, he would like to visit all the n cities one by one so that the total distance travelled is minimized. Formally, Bobo likes to find n distinct integers p1,p2,…,pn to minimize w(p1,p2)+w(p2,p3)+⋯+w(pn−1,pn) where w(x,y) is the length of road from the x-th city to the y-th city. Input The input contains at most 30 sets. For each set: The first line contains 2 integers n,m (2≤n≤105,0≤m≤105). The i-th of the following m lines contains 3 integers ai,bi,ci (1≤ai<bi≤n,1≤ci≤104). Output For each set, an integer denotes the minimum total distance. If there exists no plan, output -1 instead. Sample Input 3 3 1 2 1 1 3 1 2 3 1 3 2 1 2 1 1 3 2 Sample Output 2 -1

原题链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1245

水题：有n个城市，问按顺序从1到n的最短路径。

AC代码：

/**
  * 行有余力，则来刷题！
  * 博客链接：http://blog.csdn.net/hurmishine
  *
*/

#include 
#include 
#include 
using namespace std;
const int maxn=1e5;
const int INF=1e9;
int a[maxn];
int main()
{
    //freopen("data.txt","r",stdin);
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=2;i<=n;i++)
            a[i]=INF;
        int x,y,z;
        while(m--)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(y-x==1&&z

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