BZOJ4386[POI2015]Wycieczki / Luogu3597[POI2015]WYC - 矩乘
Solution
想到边权为$1$的情况直接矩乘就可以得出长度$<=t$ 的路径条数， 然后二分check一下即可
但是拓展到边权为$2$，$3$ 时， 需要新建节点 $i+n$ 和 $i+2n$. 从 $i+n$ 到 $i$ 连边， $i+2n$ 到 $i+n$ 连边
若 $dis[j,i]=2$，则把 $j$ 向 $i+n$连边， 距离为 $3$时同理
但是发现这样点数就有 $3*N$ 个， 二分答案+矩乘的复杂度会非常高。
那么只能用和倍增求 $LCA$ 类似的解法， 二进制枚举
复杂度为$O(N^3 logk)$
Code
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define ll long long 6 #define N 130 7 #define R register 8 using namespace std; 9 10 int n, up, m; 11 ll cnt; 12 13 int read() { 14 int X = 0, p = 1; char c = getchar(); 15 for (; c > '9' || c < '0'; c = getchar()) 16 if (c == '-') p = -1; 17 for (; c >= '0' && c <= '9'; c = getchar()) 18 X = X * 10 + c - '0'; 19 return X * p; 20 } 21 22 struct matrix { 23 ll mp[N][N]; 24 bool yue; 25 matrix() { 26 yue = false; 27 } 28 matrix operator * (const matrix &b) const { 29 matrix re; 30 memset(re.mp, 0, sizeof(re.mp)); 31 for (R int i = 0; i <= up; ++i) 32 for (R int j = 0; j <= up; ++j) { 33 for (R int k = 0; k <= up; ++k) 34 re.mp[i][j] += mp[i][k] * b.mp[k][j]; 35 if (re.mp[i][j] < 0) re.yue = true; 36 } 37 return re; 38 } 39 }ans, po[70]; 40 41 bool check(matrix tmp) { 42 ll rest = cnt; 43 if (tmp.yue) return 1; 44 for (int i = 1; i <= n; ++i) { 45 if (tmp.mp[i][0] < 0) return 1; 46 if (tmp.mp[i][0] - 1 >= rest) return 1; 47 rest -= tmp.mp[i][0] - 1; 48 } 49 return 0; 50 } 51 52 int main() 53 { 54 n = rd; m = rd; 55 up = n * 3; 56 scanf("%lld", &cnt); 57 po[0].mp[0][0] = 1; 58 for (int i = 1; i <= n; ++i) { 59 po[0].mp[i + n][i] = 1; 60 po[0].mp[i + 2 * n][i + n] = 1; 61 po[0].mp[i][0] = 1; 62 ans.mp[i][i] = 1; 63 } 64 for (R int i = 1; i <= m; ++i) { 65 int u = rd, v = rd, w = rd - 1; 66 po[0].mp[u][v + n * w]++; 67 } 68 bool flag = false; 69 int lim; 70 for (lim = 1; lim <= 65; ++lim) { 71 po[lim] = po[lim - 1] * po[lim - 1]; 72 if (check(po[lim])) { 73 flag = true; break; 74 } 75 } 76 if (!flag ) return puts("-1"), 0; 77 ll res = 0; 78 for (int i = lim; ~i; --i) { 79 matrix tmp = ans * po[i]; 80 if (!check(tmp)) ans = tmp, res += 1LL << i; 81 } 82 printf("%lld\n", res); 83 }
