Numbering Paths (Uva 125 floyd+dp思想)

Numbering Paths

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

Submit Status

Description

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 cityk'' (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 00 4 1 3 20 0 0 0 00 2 0 2 10 1 0 0 00 1 0 1 0matrix for city 10 2 1 0 0 30 0 0 0 0 10 1 0 0 0 20 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0matrix for city 2-1 -1 -1 -1 -10 0 0 0 1-1 -1 -1 -1 -1-1 -1 -1 -1 -10 0 0 0 0

题意：给出一张图，求每两个点之间的不同路径的条数，按照矩阵输出，若i到j有无数条则mp[i][j]=-1.

思路：关键是怎样判断无数条，考虑：若一个点经过一条路可以回来，即mp[i][i]>0，那么它就构成了一个环，我可以在这一无限的绕圈，知道这个了就好办了。另外一条路径经过这个点的话也是无数条。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b)  for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define mod 1000000009const int maxn = 50;const int MAXN = 2005;const int MAXM = 200010;const int N = 1005;int mp[maxn][maxn];int n,m,cas;void floyd(){    for (int k=0;k<n;k++)    {        for (int i=0;i<n;i++)        {            for (int j=0;j<n;j++)                mp[i][j]=mp[i][j]+mp[i][k]*mp[k][j];        }    }    for (int k=0;k<n;k++)    {        if (mp[k][k])        {            mp[k][k]=-1;            for (int i=0;i<n;i++)            {                for (int j=0;j<n;j++)                    if (mp[i][k]&&mp[k][j])                    mp[i][j]=-1;            }        }    }    printf("matrix for city %d\n",cas++);    for (int i=0;i<n;i++)    {        pf("%d",mp[i][0]);        for (int j=1;j<n;j++)            printf(" %d",mp[i][j]);        pf("\n");    }}int main(){#ifndef ONLINE_JUDGE    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);#endif    int i,j,u,v;    cas=0;    while (~sf(m))    {        n=0;        memset(mp,0,sizeof(mp));        for (i=0;i<m;i++)        {            sff(u,v);            mp[u][v]=1;            n=max(n,max(u,v));        }        n++;        floyd();    }    return 0;}