一、floyd算法
参考：http://blog.csdn.net/niushuai666/article/details/6772706 主要还是参考的dalao的博客理解
floyd算法用于求最短路，即两点之间最短距离，是Warshall算法的拓展。
基本思想：从节点A到节点B，只有两种可能：
1.从A直接到B
2.从A经过若干节点到B
对于第二种情况，我们假设dis(AB)为A到B的最短路，然后检查节点C，判断dis(AC)+dis(CB)<dis(AB)是否成立，若是成立，更新dis(AB) = dis(AC)+dis(CB)；在寻找AB最短的过程中，我们遍历所有的C点，对于整个图，我们遍历所有的ABC点，即三层for循环，代码如下：
for (int i = 1; i <= n; i++) {//n为节点个数 for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) { if (dis[i][k] + dis[k][j] > dis[i][j]) dis[i][j] = dis[i][k] + dis[k][j]; } } }但是我们要注意一个问题，即A的遍历顺序问题。上述代码中，i-j的最短路被过早的确定下来了。如下图
在寻找AB的时候，我们只能找到9，因为AD,DC,AC,BC的最短路并没有求出，主要是D点还没有遍历，导致相关路线也找不到最短路。
错误的主要原因是AB被过早的遍历了，由此，我们可以先遍历中间节点，然后遍历收尾节点，具体可以自己演示一遍，代码如下：
for (int k = 1; k <= n; k++) {//n为节点个数 for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (dis[i][k] + dis[k][j] > dis[i][j]) dis[i][j] = dis[i][k] + dis[k][j]; } } } 由此可见，floyd算法在节点数多的时候不占优势，时间复杂度过高，O(n3)
二、传递闭包
上个题吧，链接：http://poj.org/problem?id=3660
Cow Contest Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11232 Accepted: 6234Description
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M * Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5 4 3 4 2 3 2 1 2 2 5Sample Output
2题意：第一行表示有n个人m场比赛，后面m行每行有A B，表示A打败的B，问最后能够确定排名的人数有多少。
思路：判断一个人的排名是否确定，只要他与其余n-1个人的关系确定即可。这里涉及到了传递，如A打败了B，B打败了C，那么A间接打败了C。
这里跟floyd算法的思想是一样的，不同的是，floyd求得是最短路，if里更新的距离，这里是判断关系，不需要距离更新。
我们创建一个数字s，s[i][j]表示i打败了j，具体代码如下：
#include<iostream> #include<stdlib.h> #include<stdio.h> #include<cmath> #include<algorithm> #include<string> #include<string.h> #include<set> #include<queue> #include<stack> #include<vector> #include<functional> #include<map> using namespace std; const int maxn = 100 + 5; typedef long long LL; int s[maxn][maxn]; int n, m; void floyd() { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) { if (s[j][i] && s[i][k]) s[j][k] = 1; } } } } int main() { while (scanf("%d %d", &n, &m) != EOF) { memset(s, 0, sizeof(s)); for (int i = 1; i <= m; i++) { int a, b; scanf("%d %d", &a, &b); s[a][b] = 1; } floyd(); int sum = 0; int t = 0; for (int i = 1; i <= n; i++) { t = 0; for (int j = 1; j <= n; j++) { if (s[i][j] || s[j][i]) t++; } if (t == n - 1)sum++; } cout << sum << endl; } system("pause"); return 0; }
