XTU2016CCPC中南邀请赛C

Hamiltonian Path
Accepted : 74		Submit : 189
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

注意条件ai是小于bi！这意味着什么？意味着只能1到2,2到3的来走，任何一个跳着城市编号的道路都是无用条件，只需要把最小的相邻城市的路加起来就行了。。。mp[i]表示i到i+1的城市路径，只要条件中缺少一个mp[i]就不能满足，只能输出-1

#include <bits/stdc++.h>

typedef long long ll;
using namespace std;
const int maxn = 1e4;
int mp[maxn+10];
ll ans;

int main() {
//    freopen("in.txt","r",stdin);
    int n,m;
    int flag = 0;
    while(~scanf("%d%d",&n,&m)) {
        ans = 0LL;
        for(int i = 1; i <= maxn; i++) mp[i] = maxn;
//        cout<<ans;
        while(m--) {
            int a,b,c;
            cin>>a>>b>>c;
            if(a==b-1) {
                if(mp[a]>c) {
                    mp[a] = c;
                }
            }
        }
        for(int i = 1; i < n; i++) {
            if(mp[i]!=maxn) {
                ans += mp[i];
            }
            else {
                puts("-1");
                flag = 1;
                break;
            }
        }
        if(!flag)
            cout<<ans<<endl;
    }
}