标签:numbering paths uva 125 floyd+dp思想
Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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, 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 3There 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 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.
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.
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
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
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 1000000009 const 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; }
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Numbering Paths (Uva 125 floyd+dp思想)
标签:numbering paths uva 125 floyd+dp思想
原文地址:http://blog.csdn.net/u014422052/article/details/47282839