BZOJ 1491 社交网络(最短路)

对于这道题，如果我们能求出s到t的最短路径数目和s由v到t的最短路径数目，剩下的很好求了。

令dis[i][j]表示i到j的最短路径，dp[i][j]表示i到j的最短路径条数，如果dis[s][v]+dis[v][t]=dis[s][t]。那么必有s由v到t的最短路径条数=dp[s][v]*dp[v][t]。

我们可以进行n次dijkstra求出dis[i][j]。

对于dp数组，考虑dijkstra的过程。如果由v对u的拓展出现dis[u]==dis[v]+w,那么dp[start][u]+=dp[start][v]。

如果由v更新了节点u的最短路径，那么dp[start][u]=dp[start][v]。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

struct qnode{
    int v, c;
    qnode(int _v=0, int _c=0):v(_v),c(_c){}
    bool operator<(const qnode &r)const{return c>r.c;}
};
struct Edge{
    int v, cost;
    Edge(int _v=0, int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[N];
LL dp[N][N];
int dis[N][N], dist[N];
bool vis[N];
double ans[N];
priority_queue<qnode>que;

void Dijkstra(int n, int start){
    mem(vis,0);
    FOR(i,1,n) dist[i]=INF;
    dist[start]=0; que.push(qnode(start,0)); dp[start][start]=1;
    qnode tmp;
    while (!que.empty()) {
        tmp=que.top(); que.pop();
        int u=tmp.v;
        if (vis[u]) continue;
        vis[u]=true;
        FO(i,0,E[u].size()) {
            int v=E[u][i].v, cost=E[u][i].cost;
            if (dist[v]==dist[u]+cost) {dp[start][v]+=dp[start][u]; continue;}
            if (!vis[v]&&dist[v]>dist[u]+cost) {
                dist[v]=dist[u]+cost, que.push(qnode(v,dist[v]));
                dp[start][v]=dp[start][u];
            }
        }
    }
    FOR(i,1,n) dis[start][i]=dist[i];
}
int main ()
{
    int n, m, u, v, w;
    scanf("%d%d",&n,&m);
    FOR(i,1,m) scanf("%d%d%d",&u,&v,&w), E[u].pb(Edge(v,w)), E[v].pb(Edge(u,w));
    FOR(i,1,n) Dijkstra(n,i);
    FOR(i,1,n) FOR(j,1,n) {
        if (i==j) continue;
        FOR(k,1,n) {
            if (k==i||k==j) continue;
            if (dis[j][i]+dis[i][k]==dis[j][k]) ans[i]+=((double)dp[j][i]*dp[i][k])/(dp[j][k]);
        }
    }
    FOR(i,1,n) printf("%.3f\n",ans[i]);
    return 0;
}