最短路 & 次短路

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代码：

int dij(){    int i, j, vis[N];    for(i = 1; i <= n; i++)        vis[i] = mp[1][i];            vis[1] = 0;    for(i = 1; i <= n; i++)    {        int m = 1, f = inf;        for(j = 1; j <= n; j++)        {            if(f > vis[j] && !book[j])            {                m = j;                f = vis[j];            }        }                book[m] = 1;        for(j = 1; j <= n; j++)        {            if(!book[j] && vis[j] > vis[m] + mp[m][j])                vis[j] = vis[m] + mp[m][j];        }    }    return vis[n];}//n*log(n)struct edge {int to, cost; };typedef pair<int, int> P;  // first 是最短距离，second是顶点编号int V;vector<edge> G[Max_V];int d[Max_V];void dijkstra(int s){    //通过指定greater<P>参数，堆按照first从小到大的循序取出值    priority_queue<P, vector<P>, greater<P> > que;    fill(d, d + V, INF);    d[s] = 0;    que.push(P(0, s));        while(!que.empty())    {        P p = que.top(); que.pop();        int v = p.second;        if(d[v] < p.first) continue;        for(int i = 0; i < G[v].size(); i++)        {            edge e = G[v][i];            if(d[e.to] > d[v] + e.cost)            {                d[e.to] = d[v] + e.cost;                que.push(P(d[e.to], e.to));            }        }    }}// Bellman_Fond复杂度O(VE)struct edge {int from, to, cost; };edge es[Max_E];int d[Max_V];int V, E;void shortest_path(int s){    for(int i = 0; i < V; i++) d[i] = INF;    d[s] = 0;    while(true)    {        bool update = false;        for(int i = 0; i < E; i++)        {            edge e = es[i];            if(d[e.from != INF && d[e.to] > d[e.from] + e.cost)            {                d[e.to] = d[e.from] + e.cost;                update = true;            }        }        if(!update) break;    }}//开始时判断重边bool find_negative_loop(){    memset(d, 0, sizeof(d));    for(int i = 0; i < V; i++)    {        for(int j = 0; j < E; j++)        {            edge e = es[j];            if(d[e.to] > d[e.from] + e.cost)            {                d[e.to] = d[e.from] + e.cost;                if(i == V - 1) return false;            }        }    }    return true;}

次短路：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXN = 100009;const int INF = 0x3f3f3f3f;typedef long long LL;struct edge{    int to, cost;    edge(int tv = 0, int tc = 0):        to(tv), cost(tc) {}};typedef pair<LL,int> P;int N, R;vector<edge> graph[MAXN];LL dist[MAXN];LL dist2[MAXN];void solve(){    memset(dist, 0x3f, sizeof(dist));    memset(dist2, 0x3f, sizeof(dist2));    priority_queue<P, vector<P>, greater<P> > Q;    dist[1] = 0ll;    Q.push(P(0, 1));    while(!Q.empty())    {        P p = Q.top();        Q.pop();        int v = p.second;        LL d = p.first;        if(dist2[v] < d) continue;        for(unsigned i = 0; i < graph[v].size(); i++)        {            edge &e = graph[v][i];            LL d2 = d + e.cost;            if(dist[e.to] > d2)            {                swap(dist[e.to], d2);                Q.push(P(dist[e.to], e.to));            }            if(dist2[e.to] > d2 && dist[v] < d2)            {                dist2[e.to] = d2;                Q.push(P(dist2[e.to], e.to));            }        }    }    printf("%lld\n", dist2[N]);}int main(){    int A, B, D, t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &N, &R);        for(int i = 1; i <= N; i++) graph[i].clear();        for(int i = 0; i < R; i++)        {            scanf("%d%d%d", &A, &B, &D);            graph[A].push_back(edge(B, D));            graph[B].push_back(edge(A, D));        }        solve();    }    return 0;}

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