hdu 1625 Numbering Paths 最短路的变形，使用Floyd 外加判环

Numbering Paths

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158    Accepted Submission(s): 47

Problem Description

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.

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, j k 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: j k and k j .

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 0-1-2 and 0-2 ), 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 , 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 0-2-3-2-3-2 is a different route than 0-2-3-2 .

 

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 j k 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.

 

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 string ``matrix 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 NOT worry 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 35 0 2 0 1 1 5 2 5 2 190 1 0 2 0 30 4 1 4 2 12 03 03 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 0matrix 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 0matrix 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

 

不得不说，这题的输出格式就尼玛傻逼。PE了4,5次，每一行的第一个数就要空格。艹，最讨厌这样的题了，怪不得做的人少！

这题主要就是判环，如果mat[k][k]!=0，则K这点存在一个环，根据floyd的动态规划的含义，，mat[i][j]如果经过一个这样的k点的话，那mat[i][j]也是一个环，

直接代码：

#include <stdio.h>#include <string.h>#define MAX 1010#define INF 500000000int mat[MAX][MAX] ;void floyd(int n){for(int k = 0 ; k <= n ; ++k){for(int i = 0 ; i <= n ; ++i){if(mat[i][k] == 0)continue ;for(int j = 0 ; j <= n ; ++j){if(mat[k][j] != 0)//判环。 {if(mat[k][k] != 0 || mat[k][j] == -1 || mat[i][k] == -1){mat[i][j] = -1 ;}else{mat[i][j] += mat[i][k]*mat[k][j] ;}}}}}for(int i = 0 ; i <= n ; ++i){if(mat[i][i] > 0){mat[i][i] = -1; }}}int main(){int n ,t=0;while(scanf("%d",&n) != EOF ){int max = -INF ;memset(mat,0,sizeof(mat)) ;for(int i = 0 ; i < n ; ++i){int x , y ;scanf("%d%d",&x,&y) ;mat[x][y] = 1 ;if(x>max||y>max){max = x>y?x:y ;}}floyd(max) ;printf("matrix for city %d\n",t++);for(int i = 0; i <= max ; ++i){for(int j = 0 ; j <= max ; ++j){printf(" %d",mat[i][j]) ;}printf("\n") ;}}return 0 ;}