HDU 4650 Minimum Average Weight Path(floyd)

题意:

$N\le 100,M\le 10^4,N个点,M条边的图,无重边,可能有自环$
$所有节点对(u, v)的min\{\frac{dis(u, v)}{len(u, v)}\},不连通输出NO$

分析:

$floyd求一发连通性,判断NO$
$然后f[k][i][j]:=i到j长度为k的最短路$
$floyd的循环顺序是可以换的嘛,把i拿到最外面去枚举,这样可以复用数组$
$就变成f[k][j]:=i出发,长度为k到j的最短路$
$然后更新答案,注意由于有环,那么判断可不可以经过环$
$经过的话,环的答案如果比较小,显然可以无限绕环来逼近环长(极限思想嘛)$
$时间复杂度O(n^4)$

代码:

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//
// Created by TaoSama on 2016-03-12
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e2 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m;
int f[N][N], g[N][N], con[N][N];
double ans[N][N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &m) == 2) {
memset(g, 0x3f, sizeof g);
memset(con, 0, sizeof con);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
ans[i][j] = 1e18;
for(int i = 1; i <= m; ++i) {
int x, y, c; scanf("%d%d%d", &x, &y, &c);
g[x][y] = c;
con[x][y] = 1;
}
for(int k = 1; k <= n; ++k)
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
con[i][j] |= con[i][k] & con[k][j];
for(int i = 1; i <= n; ++i) {
memset(f, 0x3f, sizeof f);
f[0][i] = 0;
for(int l = 1; l <= n; ++l)
for(int k = 1; k <= n; ++k)
for(int j = 1; j <= n; ++j)
f[l][j] = min(f[l][j], f[l - 1][k] + g[k][j]);
for(int l = 1; l <= n; ++l)
for(int j = 1; j <= n; ++j)
ans[i][j] = min(ans[i][j], f[l][j] * 1. / l);
}
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) {
if(!con[i][j]) printf("NO%c", " \n"[j == n]);
else {
for(int k = 1; k <= n; ++k)
if(con[i][k] & con[k][j])
ans[i][j] = min(ans[i][j], ans[k][k]);
printf("%.3f%c", ans[i][j], " \n"[j == n]);
}
}
}
}
return 0;
}


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