【POJ3613】Cow Relays

Topic links: https://www.acwing.com/problem/content/347/

Title effect: Given an undirected weighted graph, to find out from \ (S \) to \ (E \) just after \ (n-\) shortest path of edges

solution

Not difficult to find, the number of vertices may be many, but not more than 100 the number of sides, so for discrete node number should first look

Matrix \ (G ^ R & lt \) , represents just after \ (R & lt \) shortest edges, wherein \ (G ^ r [i] [j] \) represents the \ (I \) to \ (J \ ) just after \ (R & lt \) shortest path edges, apparently can draw the following inferences:
\ [G ^ R & lt [I] [J] = min_ {K \} n-Leq \ left \ {G ^ P [I] [k] + G ^ q [ k] [j] \ right \} (r = p + q, n is the number of nodes) \]
also found that this formula is associative, can be quickly determined by the matrix quickly power \ (G ^ R & lt \) , \ (^ n-G [S] [E] \) is the subject of the request

code

#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &FF) {
    int RR = 1; FF = 0; char CH = getchar();
    for(; !isdigit(CH); CH = getchar()) if(CH == '-') RR = -RR;
    for(; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
    FF *= RR;
}
inline void file(string str) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 205;
int n, m, s, e, path[N][3], ni, num[N];
struct matrix{
    int g[N][N];
    matrix() {
        memset(g, 0x3f, sizeof(g));
        //for(int i = 1; i <= n; i++) g[i][i] = 0;
    }
    friend matrix operator * (const matrix &ai, const matrix &bi) {
        matrix res;
        for(int k = 1; k <= ni; k++)
            for(int i = 1; i <= ni; i++)
                for(int j = 1; j <= ni; j++)
                    res.g[i][j] = min(res.g[i][j], ai.g[i][k] + bi.g[k][j]);
        return res;
    }
}base;
void cpy(matrix &ai, const matrix &bi) {
    for(int i = 1; i <= ni; i++)
        for(int j = 1; j <= ni; j++)
            ai.g[i][j] = bi.g[i][j];
}
matrix Qpow(int ki) {
    if(ki == 1) return base;
    matrix hi = Qpow(ki / 2); cpy(hi, hi * hi);
    return ki & 1 ? hi * base : hi;
}
int main() {
    //file("");
    read(n), read(m), read(s), read(e);
    for(int i = 1; i <= m; i++) {
        read(path[i][0]), read(path[i][1]), read(path[i][2]);
        num[++ni] = path[i][1], num[++ni] = path[i][2];
    }
    sort(num + 1, num + ni + 1);
    ni = unique(num + 1, num + ni + 1) - num - 1;
    //for(int i = 1; i <= n; i++) base.g[i][i] = 0;
    for(int i = 1; i <= m; i++) {
        path[i][1] = lower_bound(num + 1, num + ni + 1, path[i][1]) - num;
        path[i][2] = lower_bound(num + 1, num + ni + 1, path[i][2]) - num;
        base.g[path[i][1]][path[i][2]] = base.g[path[i][2]][path[i][1]] = path[i][0];
    }
    s = lower_bound(num + 1, num + ni + 1, s) - num;
    e = lower_bound(num + 1, num + ni + 1, e) - num;
    matrix ans = Qpow(n);
    cout << ans.g[s][e] << endl;
    return 0;
}