POJ 2449

第K短路。

A*经典问题：

要求两点间的K短路，则图中一个点最多如队列K次（这些都是结论），

#define MAXN 10005

constint INTMAX =0x7f7f7f7f;
int minl[MAXN],vis[MAXN];
int n, m, stx, sty, k, mark;
struct  node {
    int x;
    int dis;
}u,s,v,t;
struct edge  {
    int v;
    int dis;
}p;
vector<edge> mat[10005];
vector<edge> matT[10005];
priority_queue<node> Q;

inline booloperator< (const node a,const node b) {
    return (a.dis + minl[a.x]) > (b.dis + minl[b.x]);
}


void bfs() {
    if(stx == sty) k++;
    if(minl[stx] >= INTMAX) {printf("-1\n");return ;}
    SETQ(Q);
    u.x = stx,u.dis =0;
    Q.push(u);
    while(!Q.empty()) {
        v = Q.top();Q.pop();
        //DB(v.x);
        vis[v.x]++;
        if(vis[sty] == k) {
            printf("%d\n",v.dis + minl[v.x]);
            return ;
        }
        if(vis[v.x] >k) continue;
        for (int i =0;i<mat[v.x].size() ;i++ ) {
            t.x = mat[v.x][i].v;
            t.dis = v.dis + mat[v.x][i].dis;
            Q.push(t);
        }
    }
    printf("-1\n");
}

/*void dijstra() {
    int v[MAXN],i, j, k;
    for (i=0; i<=n;i++) {
        minl[i] = INTMAX,v[i] = 0;
    }
    for (minl[sty]=0,j=0;j<n;j++) {
        for ( k =-1,i=1;i<=n ;i++ ) {
            if(!v[i]&&(k==-1||minl[i]<minl[k]))
                k=i;
        }
        for (v[k] = 1,i = 1;i<=n ;i++ ) {
            if(!v[i]&&(matt[k][i]!=-1))
                minl[i]=min(minl[k]+matt[k][i],minl[i]);
        }
    }
}*/

void spfa(int s) {
    int i;
    bool used[10005]; SET(used,false);
    for (i =0;i<10005 ;i++ ) {
        minl[i] = INTMAX;
    }
    minl[s] =0; used[s] =true;
    queue<int> q; q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop(); used[u] =false;
        for (i =0;i< matT[u].size();i++ ) {
            p = matT[u][i];
            if (minl[p.v] > minl[u] + p.dis) {
                minl[p.v] = minl[u] + p.dis;
                if (!used[p.v]) {
                    used[p.v] =true;
                    q.push(p.v);
                }
            }
        }
    }
}

void init() {
    for (int i =0;i<10005 ;i++ ) {
        mat[i].clear();
        matT[i].clear();
    }
    SET(vis,0);
    int a,b,c;
    for (int i=0; i<m;i++) {
        SCFT(a,b,c);
        p.v = b;
        p.dis = c;
        mat[a].push_back(p);
        p.v = a;
        matT[b].push_back(p);
    }
    SCFT(stx,sty,k);//DB(stx<<"--"<<sty<<"--"<<k);
    spfa(sty);
}

int main() {
    while (~SCFD(n,m)){
        init();
        bfs();
    }
}