POJ 3463 Sightseeing

求最短路，和最短路长度加1的路的条数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;const int N = 1009;const int M = 10009;const int INF =0x3f3f3f3f;int n,m,dis[N<<1],cnt[N<<1],ST,EN;bool visit[N<<1];struct node{    int to,nex,dis;} L[M];int F[N],cntF;void add(int f,int t,int dis){    L[cntF].dis = dis;    L[cntF].nex = F[f];    L[cntF].to = t;    F[f] = cntF++;}void init(){    scanf("%d%d",&n,&m);    memset(cnt,0,sizeof(cnt));    memset(F,0,sizeof(F));    cntF = 1;    int f,t,dis;    for(int i=0; i<m; i++)    {        scanf("%d%d%d",&f,&t,&dis);        add(f,t,dis);    }    scanf("%d%d",&ST,&EN);}struct nod{    int dis,to;    bool operator<(const nod t) const    {        return dis>t.dis;    }};priority_queue<nod> que;void solve(){    while(!que.empty()) que.pop();    memset(visit,false,sizeof(visit));    memset(dis,INF,sizeof(dis));    nod e,t;    e.to = ST;    e.dis = 0;    que.push(e);    dis[ST] = 0;    cnt[ST] = 1;    cnt[ST+N] = 1;    while(!que.empty())    {        e =que.top();        que.pop();        if(visit[e.to]) continue;        visit[e.to] = true;        for(int i=F[(e.to>N?e.to-N:e.to)]; i; i=L[i].nex)        {            int to = L[i].to;            int tmp=dis[e.to]+L[i].dis;            if(dis[to]>tmp)            {                if(dis[to]!=INF)                {                    dis[to+N] = dis[to];                    cnt[to+N] = cnt[to];                    t.dis=dis[to+N];                    t.to=to+N;                    que.push(t);                }                dis[to] = tmp;                cnt[to] = cnt[e.to];                t.dis = dis[to];                t.to = to;                que.push(t);            }            else if(dis[to] == tmp)            {                cnt[to] += cnt[e.to];            }            else if(dis[to+N] > tmp)            {                dis[to+N] = tmp;                cnt[to+N] = cnt[e.to];                t.dis = dis[to+N];                t.to = to+N;                que.push(t);            }            else if(dis[to+N] == tmp)            {                cnt[to+N] += cnt[e.to];            }        }    }    printf("%d\n",cnt[EN]+(dis[EN+N]-dis[EN]==1?cnt[EN+N]:0));}int main(){    freopen("in.txt","r",stdin);    int cas;    scanf("%d",&cas);    while(cas--)    {        init();        solve();    }    return 0;}