UVA - 821 - Page Hopping（floyd算法）

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题意：用标号给定几个网站，一个标号（1~100）代表一个网站（标号的给定无连续性）。可以通过一个网站访问另一个网站（通过给定的有向图路径），题目保证任意两个网站都有路到达且输入无自环，求每两个网站访问，最少需要经过几个网站的个数平均值。

先重整标号使其有序，再用floyd算法求出每两个的最短路，最后统计即可。

#include<cstdio>#include<cstring>#include<cctype>#include<cstdlib>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<list>typedef long long ll;typedef unsigned long long llu;const int MAXN = 100 + 10;const int MAXT = 10000 + 10;const int INF = 0x7f7f7f7f;const double pi = acos(-1.0);const double EPS = 1e-6;using namespace std;int lur;double g[MAXN][MAXN];map<int, int> mp;int getid(int x){    if(mp.count(x))  return mp[x];    return mp[x] = ++lur;}void floyd(){    for(int k = 1; k <= lur; ++k)        for(int i = 1; i <= lur; ++i)            for(int j = 1; j <= lur; ++j)                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);}void merge(int a, int b){    int x = getid(a), y = getid(b);    g[x][y] = 1;}double solve(){    floyd();    double sum = 0, num = 0;    for(int i = 1; i <= lur; ++i)        for(int j = 1; j <= lur; ++j){            if(i == j)  continue;            num += 1, sum += g[i][j];        }    return sum / num;}int main(){    int a, b, cas = 0;    while(scanf("%d%d", &a, &b) == 2 && (a || b)){        for(int i = 0; i < MAXN; ++i)  for(int j = 0; j < MAXN; ++j)  g[i][j] = 1e9;        mp.clear();        lur = 0;        merge(a, b);        while(scanf("%d%d", &a, &b) == 2 && (a || b))  merge(a, b);        printf("Case %d: average length between pages = %.3lf clicks\n", ++cas, solve());    }    return 0;}

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