次短路(hdu 6181)

题意:求次短路

For testcase 1, Alice take path 1 - 3 and its length is 3, and then Bob will take path 1 - 2 - 3 and its length is 5.For testcase 2, Bob will take route 1 - 2 - 1 - 2 and its length is 3

次短路在dijkstra的时候 记录次小路径，具体看代码

#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
#define maxn 100050
#define INF (1LL<<60)
typedef long long int ll;
typedef pair<long long int,int> P;
int T;
int n , m;

ll dis[maxn];
ll disc[maxn];
vector<P>E[maxn];

void dj()
{
ll mi = INF;
for(int i = 0 ; i <= n ; ++i)dis[i] = disc[i] = INF;
dis[1] = 0;//只能赋值最短路的起始点为 0 ，否则次短路无法更新到原点
priority_queue<P,vector<P>,greater<P> > Q;
Q.push(P(0,1));
while(! Q.empty())
{
ll c = Q.top().first;
int u = Q.top().second;
Q.pop();
if( disc[u] < c )continue;
for(int i = 0 ; i < E[u].size() ; ++i)
{
int v = E[u][i].first;
ll w = E[u][i].second;

ll tmp = w + c;
if( dis[v] >= tmp )
{
swap(tmp,dis[v]);//此处注意交换
Q.push(P(dis[v],v));
}
if( disc[v] > tmp && dis[v] < tmp )
{
disc[v] = tmp;
Q.push( P(tmp,v) ) ;
}
}
}

printf("%lld\n",disc
);
}

int main()
{
cin >> T;
while( T-- )
{
cin >> n >> m;
for(int i = 0 ; i <= n ; ++i)E[i].clear();
for(int i = 0 ; i < m ; ++i)
{
int a,b;
ll c;
scanf("%d %d %lld",&a,&b,&c);
E[a].push_back(P(b,c));
E[b].push_back(P(a,c));
}

dj();

}
}