HDU 2807 The Shortest Path 矩阵 + Floyd

水题， 直接模拟就可以。姿势要注意。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 81;

struct Matrix {
    int n, m, data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }

    bool operator == (const Matrix &x) const {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                if(data[i][j] != x.data[i][j])
                    return false;
            }
        }
        return true;
    }

    void print() {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                printf("%d ", data[i][j]);
            }
            puts("");
        }
    }
};

Matrix operator * (Matrix &a, Matrix &b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
            }
        }
    }
    return ret;
}

Matrix city[maxn];
int n, m, dist[maxn][maxn];

int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        if(n == 0 && m == 0) break;
        for(int i = 1; i <= n; i++) {
            city[i].n = city[i].m = m;
            for(int j = 1; j <= m; j++) {
                for(int k = 1; k <= m; k++) {
                    scanf("%d", &city[i].data[j][k]);
                }
            }
        }
        memset(dist, 0, sizeof(dist));
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) dist[i][j] = INF;
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) if(i != j) {
                Matrix now = city[i] * city[j];
                for(int k = 1; k <= n; k++) if(k != j && k != i) {
                    if(now == city[k]) {
                        dist[i][k] = 1;
                    }
                }
            }
        }
        //floyd
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                for(int k = 1; k <= n; k++) {
                    if(dist[j][i] + dist[i][k] < dist[j][k]) {
                        dist[j][k] = dist[j][i] + dist[i][k];
                    }
                }
            }
        }
        int q; scanf("%d", &q);
        while(q--) {
            int x, y; scanf("%d%d", &x, &y);
            if(dist[x][y] < INF) printf("%d\n", dist[x][y]);
            else puts("Sorry");
        }
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/rolight/p/4050089.html