HDU

题意：求次短路。

分析：关键是情况讨论。

LL tmpd = x.d + e.dist;

以下情况对应的更新结果

1、tmpd（2） < 最短路（3） < 次短路（4）-------> 最短路 = 2，次短路 = 3

2、tmpd（2） = 最短路（2） < 次短路（3）-------> 最短路 = 2，次短路 = 2

3、最短路（2） < tmpd（3） < 次短路（4）-------> 最短路 = 2，次短路 = 3

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Edge{
    int from, to;
    LL dist;
    Edge(int f, int t, LL d):from(f), to(t), dist(d){}
};
struct HeapNode{
    LL d;
    int u;
    HeapNode(LL dd, int uu):d(dd), u(uu){}
    bool operator < (const HeapNode& rhs)const{
        return d > rhs.d;
    }
};
struct Dijkstra{
    int n, m;
    vector<Edge> edges;
    vector<int> G[MAXN];
    LL dist1[MAXN];
    LL dist2[MAXN];
    void init(int n){
        this -> n = n;
        for(int i = 0; i < n; ++i) G[i].clear();
        edges.clear();
    }
    void AddEdge(int from, int to, LL dist){
        edges.push_back(Edge(from, to, dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }
    void dijkstra(int s){
        priority_queue<HeapNode> Q;
        for(int i = 0; i < n; ++i){
            dist1[i] = LL_INF;
            dist2[i] = LL_INF;
        }
        dist1[s] = 0;
        Q.push(HeapNode(0, s));
        while(!Q.empty()){
            HeapNode x = Q.top();
            Q.pop();
            int u = x.u;
            if(x.d > dist2[u]) continue;
            for(int i = 0; i < G[u].size(); ++i) {
                Edge &e = edges[G[u][i]];
                LL tmpd = x.d + e.dist;
                if(tmpd < dist1[e.to]){
                    swap(tmpd, dist1[e.to]);
                    Q.push(HeapNode(dist1[e.to], e.to));
                }
                else if(tmpd == dist1[e.to]){
                    dist2[e.to] = tmpd;
                    Q.push(HeapNode(dist2[e.to], e.to));
                }
                if(tmpd > dist1[e.to] && tmpd < dist2[e.to]){
                    dist2[e.to] = tmpd;
                    Q.push(HeapNode(dist2[e.to], e.to));
                }
            }
        }
    }
}dij;
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n, m;
        scanf("%d%d", &n, &m);
        dij.init(n);
        int x, y;
        LL d;
        for(int i = 0; i < m; ++i){
            scanf("%d%d%lld", &x, &y, &d);
            dij.AddEdge(x - 1, y - 1, d);
            dij.AddEdge(y - 1, x - 1, d);
        }
        dij.dijkstra(0);
        printf("%lld
", dij.dist2[n - 1]);
    }
    return 0;
}