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

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rd read()
 #define ll long long
 #define N 130
 #define R register
 using namespace std;

 int n, up, m;
 ll cnt;

 int read() {
     int X = , p = ; char c = getchar();
     for (; c > '' || c < ''; c = getchar())
         if (c == '-') p = -;
     for (; c >= '' && c <= ''; c = getchar())
         X = X *  + c - '';
     return X * p;
 }

 struct matrix {
     ll mp[N][N];
     bool yue;
     matrix() {
         yue = false;
     }
     matrix operator * (const matrix &b) const {
         matrix re;
         memset(re.mp, , sizeof(re.mp));
         for (R int i = ; i <= up; ++i)
             for (R int j = ; j <= up; ++j) {
                 for (R int k = ; k <= up; ++k)
                     re.mp[i][j] += mp[i][k] * b.mp[k][j];
                 if (re.mp[i][j] < ) re.yue = true;
             }
         return re;
     }
 }ans, po[];

 bool check(matrix tmp) {
     ll rest = cnt;
     if (tmp.yue) return ;
     for (int i = ; i <= n; ++i) {
         if (tmp.mp[i][] < ) return ;
         if (tmp.mp[i][] -  >= rest) return ;
         rest -= tmp.mp[i][] - ;
     }
     return ;
 }

 int main()
 {
     n = rd; m = rd;
     up = n * ;
     scanf("%lld", &cnt);
     po[].mp[][] = ;
     for (int i = ; i <= n; ++i) {
         po[].mp[i + n][i] = ;
         po[].mp[i +  * n][i + n] = ;
         po[].mp[i][] = ;
         ans.mp[i][i] = ;
     }
     for (R int i = ; i <= m; ++i) {
         int u = rd, v = rd, w = rd - ;
         po[].mp[u][v + n * w]++;
     }
     bool flag = false;
     int lim;
     for (lim = ; lim <= ; ++lim) {
         po[lim] = po[lim - ] * po[lim - ];
         if (check(po[lim])) {
             flag = true; break;
         }
     }
     if (!flag ) return puts("-1"), ;
     ll res = ;
     for (int i = lim; ~i; --i) {
         matrix tmp = ans * po[i];
         if (!check(tmp)) ans = tmp, res += 1LL << i;
     }
     printf("%lld\n", res);
 }