uva 125 Numbering Paths

原题：
Background
Problems that process input and generate a simple ‘yes’ or ‘no’ answer are called decision problems. One class of decision problems, the NP-complete problems, are not amenable to general efficient solutions. Other problems may be simple as decision problems, but enumerating all possible ‘yes’ answers may be very difficult (or at least time-consuming).
This problem involves determining the number of routes available to an emergency vehicle operating in a city of one-way streets.
The Problem
Given the intersections connected by one-way streets in a city, you are to write a program that determines the number of different routes between each intersection. A route is a sequence of one-way streets connecting two intersections.
Intersections are identified by non-negative integers. A one-way street is specified by a pair of intersections. For example, tex2html_wrap_inline30 indicates that there is a one-way street from intersection j to intersection k. Note that two-way streets can be modeled by specifying two one-way streets: tex2html_wrap_inline30 and tex2html_wrap_inline38 .
Consider a city of four intersections connected by the following one-way streets:
0 1
0 2
1 2
2 3
There is one route from intersection 0 to 1, two routes from 0 to 2 (the routes are tex2html_wrap_inline40 and tex2html_wrap_inline42 ), two routes from 0 to 3, one route from 1 to 2, one route from 1 to 3, one route from 2 to 3, and no other routes.
It is possible for an infinite number of different routes to exist. For example if the intersections above are augmented by the street tex2html_wrap_inline44 , there is still only one route from 0 to 1, but there are infinitely many different routes from 0 to 2. This is because the street from 2 to 3 and back to 2 can be repeated yielding a different sequence of streets and hence a different route. Thus the route tex2html_wrap_inline46 is a different route than tex2html_wrap_inline48 .
The Input
The input is a sequence of city specifications. Each specification begins with the number of one-way streets in the city followed by that many one-way streets given as pairs of intersections. Each pair tex2html_wrap_inline30 represents a one-way street from intersection j to intersection k. In all cities, intersections are numbered sequentially from 0 to the largest'' intersection. All integers in the input are separated by whitespace. The input is terminated by end-of-file.
There will never be a one-way street from an intersection to itself. No city will have more than 30 intersections.
The Output
For each city specification, a square matrix of the number of different routes from intersection j to intersection k is printed. If the matrix is denoted M, then M[j][k] is the number of different routes from intersection j to intersection k. The matrix M should be printed in row-major order, one row per line. Each matrix should be preceded by the stringmatrix for city k” (with k appropriately instantiated, beginning with 0).
If there are an infinite number of different paths between two intersections a -1 should be printed. DO NOTworry about justifying and aligning the output of each matrix. All entries in a row should be separated by whitespace.
Sample Input
7 0 1 0 2 0 4 2 4 2 3 3 1 4 3
5
0 2
0 1 1 5 2 5 2 1
9
0 1 0 2 0 3
0 4 1 4 2 1
2 0
3 0
3 1
Sample Output
matrix for city 0
0 4 1 3 2
0 0 0 0 0
0 2 0 2 1
0 1 0 0 0
0 1 0 1 0
matrix for city 1
0 2 1 0 0 3
0 0 0 0 0 1
0 1 0 0 0 2
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
matrix for city 2
-1 -1 -1 -1 -1
0 0 0 0 1
-1 -1 -1 -1 -1
-1 -1 -1 -1 -1
0 0 0 0 0

中文：
给你一个有向图，然后让你输出一个矩阵，表示出从一个点到另外一个点的路径有多少条，如果有无数条，那么这两个点之间用-1表示。

#include<bits/stdc++.h>
using namespace std;
int n,ind;
int mat[31][31];

int main()
{
    ios::sync_with_stdio(false);
 //   fstream in,out;
    int t=0;
//    in.open("in.txt");
//    out.open("out.txt");
    while(cin>>n)
    {
        memset(mat,0,sizeof(mat));
        ind=0;
        for(int i=0;i<n;i++)
        {
            int a,b;
            cin>>a>>b;
            mat[a][b]=1;
            ind=max(ind,max(a,b));

        }
        for(int k=0;k<=ind;k++)
        {
            for(int i=0;i<=ind;i++)
            {
                for(int j=0;j<=ind;j++)
                {
                    mat[i][j]+=mat[i][k]*mat[k][j];
                }
            }
        }
        for(int k=0;k<=ind;k++)
        {
            if(mat[k][k]!=0)
            for(int i=0;i<=ind;i++)
            {
                for(int j=0;j<=ind;j++)
                {
                    if(mat[i][k]!=0&&mat[k][j]!=0)
                        mat[i][j]=-1;
                }
            }
        }
        cout<<"matrix for city "<<t++<<endl;
        for(int i=0;i<=ind;i++)
        {
            for(int j=0;j<=ind;j++)
            {
                cout<<mat[i][j];
                if(j==ind)
                    cout<<endl;
                else
                    cout<<" ";
            }
        }
    }
 //   in.close();
 //   out.close();
    return 0;
}

思路：

第一眼看到这题就立刻就想起Matrix 67的一篇文章，就是讲矩阵的那部分。
原文链接http://www.matrix67.com/blog/archives/276
所以很快就把代码写出来，结果一直wa，上网上搜了搜其他人的答案，思路也是大同小异，只不过是用floyed算法改进而来。

基本上的思路是这样的，mat[i][j]表示从i到j的路径数，那么mat[i][j]=sigma mat[i][jk]*mat[k][j]
这里k只中间节点