codeforces Round #241(div2) E解题报告

E. President's Path time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Good old Berland has n cities and m roads. Each road connects a pair of distinct cities and is bidirectional.

E. President's Path

time limit per test

memory limit per test

input

output

Good old Berland hasncities andmroads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.

We also know that the President will soon ride along the Berland roads from citysto cityt. Naturally, he will choose one of the shortest paths fromstot, but nobody can say for sure which path he will choose.

The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.

Making the budget for such an event is not an easy task. For all possible distinct pairss,?t(s?t) find the number of roads that lie on at least one shortest path fromstot.

Input

The first line of the input contains integersn,?m(2?≤?n?≤?500,0?≤?m?≤?n·(n?-?1)?/?2) — the number of cities and roads, correspondingly. Thenmlines follow, containing the road descriptions, one description per line. Each description contains three integersxi,?yi,?li(1?≤?xi,?yi?≤?n,?xi?≠?yi,?1?≤?li?≤?106), wherexi,?yiare the numbers of the cities connected by thei-th road andliis its length.

Output

Print the sequence ofintegersc12,?c13,?...,?c1n,?c23,?c24,?...,?c2n,?...,?cn?-?1,?n, wherecstis the number of roads that can lie on the shortest path fromstot. Print the elements of sequencecin the described order. If the pair of citiessandtdon't have a path between them, thencst?=?0.

Sample test(s)

input

5 6 1 2 1 2 3 1 3 4 1 4 1 1 2 4 2 4 5 4

output

1 4 1 2 1 5 6 1 2 1

题目大意:

给出一个图,要求求出每个点之间的最短距离的路总条数.

解法:

由于需要求出每个点之间的最短距离的路总条数,我们可以将问题拆分为: 1.最短距离; 2.符合最短距离的路的总条数.

对于问题1,可以用floyd算法可以算出来;

对于问题2,本题的重点所在,要求出路的总条数,而且不能重复.不能重复的话,我们可以按照每个点来计算,且每个点,我们只计算点周围的最短路径的路的条数,这样就可以防止重复. 因为1条边不可能统计2次.(详见代码).

代码:

#include  #include  #include  #include  using namespace std; const int lim = 500000001; class TMain { private: int n, m, f[505][505], dis[505][505], ans[505][505], now[505]; public: int run() { scanf("%d %d", &n, &m); for (int i = 1; i