hdu1625 Numbering Paths (floyd判环)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 207    Accepted Submission(s): 63

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

 

题意：给你n条边，让你求出一个点到另一个点能走的路线的条数，如果能走的路线中有环，那么就输出-1.

思路：因为数据很小，只有30，所以可以用floyd先算出每两个点之间的路线条数，方法为f[i][j]+=gra[i][k]*gra[k][j].然后循环每一个点，判断f[i][i]是不是为0，如果不为0，那么说明i这个点在环上，之后只要看任意两点j能不能经过i后到达k，如果能，那么f[j][k]就是-1.

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef unsigned long long ll;#define inf 99999999#define pi acos(-1.0)#define maxn 505#define maxnode 100int gra[40][40];int main(){    int n,m,i,j,c,d,k;    int cas=0;    while(scanf("%d",&m)!=EOF)    {        n=0;        memset(gra,0,sizeof(gra));        for(i=1;i<=m;i++){            scanf("%d%d",&c,&d);            gra[c][d]=1;            n=max(n,c);            n=max(n,d);        }        for(k=0;k<=n;k++){            for(i=0;i<=n;i++){                for(j=0;j<=n;j++){                    gra[i][j]+=gra[i][k]*gra[k][j];                }            }        }        for(i=0;i<=n;i++){            if(gra[i][i]){                gra[i][i]=-1;                for(j=0;j<=n;j++){                    for(k=0;k<=n;k++){                        if(gra[j][i] && gra[i][k]){                            gra[j][k]=-1;                        }                    }                }            }        }        printf("matrix for city %d\n",cas++);        for(i=0;i<=n;i++){            for(j=0;j<=n;j++){                printf(" %d",gra[i][j]);            }            printf("\n");        }    }}