矩阵数字的一个路径的算法，怎么利用C语言编写程序的方式实现的

Question

Problem Description
Problems that process input and generate a simple yes'' orno'' 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 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

Answer 1

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#include <stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    while(n--) {
        int m[31][31];
        for(int i = 0; i <= 30; ++i) {
            for(int j = 0; j <= 30; ++j)
                m[i][j] = -1;
        }
        int a, b;
        scanf("%d", &a);
        while(a--) {
            scanf("%d %d", &b, &c);
            m[b][c] = 0;
        }
        for(int i = 0; i <= 30; ++i) {
            for(int j = 0; j <= 30; ++j)
                printf("%d ", m[i][j]);
            printf("\n");
        }
        printf("matrix for city %d\n", i + 1);
    }
    return 0;
}