HDU 6181 Two Paths

第一次写次短路

从起点和终点出发分别跑两次单源最短路得到dis1[] 和 dis2[]，对所有边e的起点st，终点ed，将备选方案dis1[st]+dis2[ed]+e.length加入ans中，ans排序，选取次小的结果就是答案

没看数据范围错了两次，下次注意……

#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
vector<pair<LL,LL> > v[100100];
int vis[100100];
int st,ed,ind,test,n,m;
LL dis,len;
pair<LL,LL> tmp;
LL dis1[100100],dis2[100100];
priority_queue<pair<LL,LL>,vector<pair<LL,LL> >,greater<pair<LL,LL> > > q;
void dij(int st,LL *arr){
	while(!q.empty()) q.pop();
	memset(vis,0,sizeof(vis));
	for(int i = 1;i<=n;i++) arr[i] = 1e17;
	arr[st] = 0;
	q.push(make_pair(0,st));
	while(!q.empty()){
		tmp = q.top();
		q.pop();
		ind = tmp.second;
		if(vis[ind]) continue;
		vis[ind] = 1;
		arr[ind] = dis = tmp.first;
		for(int i = 0;i<v[ind].size();i++){
			int to = v[ind][i].second;
			if(!vis[to]){
				if(arr[to]>arr[ind]+v[ind][i].first) arr[to] = arr[ind]+v[ind][i].first,q.push(make_pair(arr[to],to));
			}
		}
	}
}
int main(){
	scanf("%d",&test);
	while(test--){
		scanf("%d%d",&n,&m);
		for(int i = 1;i<=n;i++) v[i].clear();
		for(int i = 0;i<m;i++){
			scanf("%d%d%lld",&st,&ed,&len);
			v[st].push_back(make_pair(len,ed));
			v[ed].push_back(make_pair(len,st));
		}
		dij(1,dis1);
		dij(n,dis2);
		//for(int i = 1;i<=n;i++) cout<<"dis1: "<<i<<" "<<dis1[i]<<endl;
		//for(int i = 1;i<=n;i++) cout<<"dis2: "<<i<<" "<<dis2[i]<<endl;
		vector<LL> ans;
		for(int i = 1;i<=n;i++){
			for(int j = 0;j<v[i].size();j++){
				int to = v[i][j].second;
				ans.push_back(v[i][j].first+dis1[i]+dis2[to]);
			}
		}
		sort(ans.begin(),ans.end());
		LL mmin = ans[0];
		for(int i = 0;i<ans.size();i++){
			if(ans[i] != mmin){
				cout<<ans[i]<<endl;
				break;
			}
		}
	}
	return 0;
}