Numbering Paths

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, 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: and .

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 and ), 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 is a different route than .

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 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 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

---

做法 : floyd-warshall

#include <stdio.h>
#include <string.h>
int cnt[31][31] = {};
int m, n, test = 0, x, y;
int main() {
    int i, j, k;
    while(scanf("%d", &m) == 1) {
        memset(cnt, 0, sizeof(cnt));
        n = 0;
        while(m--) {
            scanf("%d %d", &x, &y);
            cnt[x][y]++;
            if(x > n) n = x;
            if(y > n) n = y;
        }
        for(k = 0; k <= n; k++) {
            for(i = 0; i <= n; i++) {
                for(j = 0; j <= n; j++) {
                    if(cnt[i][k] != 0 && cnt[k][j] != 0) {
                        cnt[i][j] += cnt[i][k]*cnt[k][j];
                    }
                }
            }
        }
        for(k = 0; k <= n; k++) {
            if(cnt[k][k])
            for(i = 0; i <= n; i++) {
                for(j = 0; j <= n; j++) {
                    if(cnt[i][k] != 0 && cnt[k][j] != 0) {
                        cnt[i][j] = -1;
                    }
                }
            }
        }
        printf("matrix for city %d\n", test++);
        for(i = 0; i <= n; i++) {
            for(j = 0; j <= n; j++) {
                if(j)
                    putchar(' ');
                printf("%d", cnt[i][j]);
            }
            puts("");
        }
    }
    return 0;
}