POJ3255（次小生成树）

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次小生成树裸题

到某个点的次小次短路要么是其它某个点u的最短路加上u->v的边，要么就是到u的次短路加上u->v的边，因此所需要求的就是到所有顶点的最短路和次短路

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int maxn = 5000 + 10;const int INF = 1<<30;typedef pair<int, int> P;int N, R;struct edge{    int to, cost;};vector<edge> G[maxn];int dist[maxn];  //最短路径int dist2[maxn];  //次短路经void solve(){    priority_queue<P, vector<P>, greater<P> > que;    fill(dist, dist+N, INF);  //最短距离    fill(dist2, dist2+N, INF); //次短距离    dist[0] = 0;    que.push(P(0,0));    while(!que.empty())    {        P p = que.top();        que.pop();        int v = p.second, d = p.first;        if(dist2[v] < d) continue;        for(int i=0; i<G[v].size(); ++i)        {            edge e = G[v][i];            int d2 = d + e.cost;            if(dist[e.to]>d2)            {                swap(dist[e.to], d2);                que.push(P(dist[e.to], e.to));            }            if(dist2[e.to] >d2&&dist[e.to] <d2)            {                dist2[e.to] = d2;                que.push(P(dist2[e.to], e.to));            }        }    }    printf("%d\n",dist2[N-1]);}int main(){    int i, a, b, c;    edge e;    while(~scanf("%d%d",&N,&R))    {        for(i=0; i<R; ++i)        {            scanf("%d%d%d", &a, &b, &c);            a--,b--;            e.to = b, e.cost = c;            G[a].push_back(e);            e.to = a;            G[b].push_back(e);        }        solve();    }    return 0;}

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