# 1482 Route Statistics

Source: Internet
Author: User

1482 Route Statistics

time limit: 1 sspace limit: 256000 KBtitle level: Diamonds Diamond SolvingView Run ResultsTitle Description Description

A graph of n nodes, which is the total number of scenarios from start to finish just over time. MoD 502630.

Enter a description Input Description

The first line contains an integer n, and all points are numbered from 0 to N-1.

Next n rows, each row contains n characters. The J-Character of line I indicates the time required for I to J. Characters can only be ' 1 ' to ' 5 ', or '. ' Indicates that I cannot reach J. Ensure that the main diagonal is '. '.

Next line 3 integers start, finish, time.

Output description Output Description

The total number of output scenarios.

Sample input Sample Input

3

.12

2.1

12.

0 2 5

Sample output Sample Output

8

Data range and Tips Data Size & Hint

For 20% of the data, the character entered is not ' 1 ' is '. ';

For 100% data, 1 <= n <= 10; 1 <= start,finish <= N; 1 <= time <= 10^9.

"Problem Solving Report"

First glance, DFS, but look at the range of T, apparently timed out.

Again, the point is very small (the matrix of the N power time-consuming less), a lot of time (go complex)

The symbol of matrix multiplication Ah!!!

We know the infinity graph from S through K steps to f how to ask: that is 01 matrix:

Establish matrix A when and only if there is an edge i->j, a (i,j) = 1.

So c=a*a, then C (i,j) =σa (i,k) *a (k,j), is actually equal to the number of paths from point I to J that passes exactly 2 edges (enumeration K is the transit point)

Similarly, the first row J of C*a indicates the number of paths from I to J through 3 edges.

Similarly, if the number of paths through the K-step is required, we only need two points to find out the a^k.

But the figure has the edge, how to do?

Two characters: split!

The relationship between each point is stored in a matrix, I can be 1 steps to J Mark 1, not to mark 0, note the right side of the title is 1-5, then can be split, each point is split into a point of edge,

But that's not enough, we're going to build (n*n*5) ^2 = 500*500 matrix, matrix multiplication T reaches 500 ^ 3 This is obviously too much.

So: We encountered a side i,j, the right to C, it was split into i–> i+n*1-i+n*2->, ..... i+n* (C-1), J

:

So there is only (n*5) the matrix of ^2=50*50.

The matrix becomes the 01 matrix when the point is removed

The a[i][j in this moment] saves 1 steps from I to J, requiring T-step, then directly squared the matrix to the T-time answer.

AC Code:

`#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespacestd;Const intMod=502630;intF,t,s,n;CharC;intans;intCNT;structnode{intf[ -][ -];} e,x;voidClean () { for(intI=0; i<=n;i++) e.f[i][i]=1;} Node Cheng (node A,node b) {//matrix multiplicationnode NE;  for(intI=0; i<n;i++){         for(intj=0; j<n;j++) {Ne.f[i][j]=0;  for(intk=0; k<n;k++) ne.f[i][j]= (ne.f[i][j]+ (Long Long) a.f[i][k]*b.f[k][j])%mod)%MoD; }    }    returnNE;}intAnswer () {node NE=x; Node=D; intb=T;  while(b) {//Matrix exponentiation        if(b&1) ass=Cheng (ass,ne); b>>=1; NE=Cheng (ne,ne); }    returnass.f[s][f];}intMain () {scanf ("%d\n",&N); CNT=N;  for(intI=0; i<n;i++){         for(intj=0; j<n;j++) {scanf ("%c",&c); if(c>='0'&&c<='9') {CNT=c-'0';  for(intk=1; k<cnt;k++) x.f[n* (K-1) +i][n*k+i]=1;//Split Pointx.f[n* (cnt-1) +i][j]=1; }} scanf ("\ n"); } N*=5;//last N to multiply by 5scanf"%d%d%d",&s,&f,&t);                    Clean (); //Defining the Unit matrixprintf"%d\n", Answer ()); return 0;}`

1482 Route Statistics

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