(2种状态最短路)nuboj 2473

题目链接：http://www.nbuoj.com/v8.8/Problems/Problem.php?pid=2473

题意：中文题

题解：见代码，注意初始建边。

#include 

using namespace std;

#define pb push_back

#define Pii  pair  // first dis ,second u

const int N = 2E5 + 7;
vectorG[N];
bool a[N], vis[N];
int n, m;
int dis[N];

void add(int u, int v, int w, bool sta) //规定到达u点时，接下来u要减速为不正常点,否则正常
{
    if(sta) {
        G[u].pb(Pii(2 * w, n + v)); //u->v 建立一条2c的边, 并且到达v点时不正常
        G[u + n].pb(Pii(2 * w, n + v)); //也可以由不正常u到达
    } else {
        G[u].pb(Pii(w, v)); //常规建边
        G[u + n].pb(Pii(2 * w, v)); //不正常u点经过了一次减速，在减速一次到达正常v点
    }
}

void dij()
{
    priority_queue, greater > que;
    que.push(Pii(0,0));
    while(!que.empty()) {
        int u = que.top().second;
        que.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = 0;i < G[u].size();i ++) {
            Pii t = G[u][i];
            if(!vis[t.second] && dis[t.second] > t.first + dis[u]) {
                dis[t.second] = t.first + dis[u];
                que.push(Pii(dis[t.second], t.second));
            }
        }
    }
}
int main()
{

    scanf("%d %d", &n, &m);

    memset(dis, 0x3f, sizeof(dis));
    dis[0] = 0;
    dis[n] = 0;
    for(int i = 0;i < n;i ++) scanf("%d", &a[i]);
    for(int i = 1;i <= m;i ++) {
        int u, v, w;
        scanf("%d %d %d", &u, &v, &w);
        add(u, v, w, a[u]);
        add(v, u, w, a[v]);
    }
    dij();
    printf("%d\n", min(dis[n-1], dis[2*n-1]));
    return 0;
}