fzuoj 2173(矩阵快速幂)
思路:用邻接矩阵存储图,然后矩阵的k次方即为答案。只需要修改矩阵乘法c[i][j] = min(c[i][j], a[i][k] + b[k][j])即可。并不难写关键是思路。
代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 #include <set> 8 #include <map> 9 #include <string> 10 #include <math.h> 11 #include <stdlib.h> 12 #include <time.h> 13 using namespace std; 14 15 const int LEN = 51; 16 typedef long long ll; 17 const ll LINF = 0x3f3f3f3f3f3f3f3fLL; 18 int n, h, k; 19 20 21 void debug(ll Mix[][LEN]){ 22 for(int i=0; i<n; i++){ 23 for(int j=0; j<n; j++){ 24 if(Mix[i][j] != LINF)cout << Mix[i][j] << ' ' ; 25 else cout << "- "; 26 } 27 cout << endl; 28 }cout << endl; 29 } 30 31 void Mul(ll a[][LEN], ll b[][LEN]){ 32 ll c[LEN][LEN]; 33 memset(c, 0x3f, sizeof c); 34 for(int i=0; i<n; i++){ 35 for(int j=0; j<n; j++){ 36 for(int k=0; k<n; k++){ 37 if(a[i][k] == LINF || b[k][j] == LINF) continue; 38 c[i][j] = min(c[i][j], a[i][k] + b[k][j]); 39 } 40 } 41 } 42 // debug(a); debug(b);debug(c); 43 for(int i=0; i<n; i++){ 44 for(int j=0; j<n; j++){ 45 a[i][j] = c[i][j]; 46 } 47 } 48 } 49 50 void Mksm(ll t[][LEN], int k){ 51 ll Mix[LEN][LEN]; 52 for(int i=0; i<n; i++) 53 for(int j=0; j<n; j++) 54 Mix[i][j] = (i==j ? 0 : LINF); 55 while(k){ 56 if(k & 1) Mul(Mix, t); 57 Mul(t, t); 58 k >>= 1; 59 } 60 for(int i=0; i<n; i++) 61 for(int j=0; j<n; j++) 62 t[i][j] = Mix[i][j]; 63 } 64 65 66 int main() 67 { 68 // freopen("in.txt","r",stdin); 69 //freopen("out.txt","w",stdout); 70 71 int T, a, b, val; 72 ll Mix[LEN][LEN]; 73 cin >> T; 74 while(T--){ 75 memset(Mix, 0x3f, sizeof Mix); 76 cin >> n >> h >> k; 77 for(int i=0; i<h; i++){ 78 cin >> a >> b >> val; 79 a--, b--; 80 Mix[a][b] = val; 81 } 82 Mksm(Mix, k); 83 if(Mix[0][n-1] != LINF) cout << Mix[0][n-1] << endl; 84 else cout << -1 << endl; 85 } 86 return 0; 87 }
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