hdu 2807 矩阵运算+最短路

题意为有N个城市，每个城市用一个矩阵表示，如果三个矩阵A、B、C满足A*B=C，则A到C有一条长度为1的路径，问两个城市间的最短路。这题的最短路部分很简单，稍微麻烦的是矩阵处理。想了一下，觉得优化的矩阵运算不好打，就直接打了个最简单的，AC了，可见杭电OJ的测试数据还是很弱的。这题也WA了一次，原因是以为A*B=B，则A到B也有一条路径，测试表明测试数据中这种情况认为A到B是没有路径的。

/*
 * hdu2807/win.c
 * Created on: 2011-7-23
 * Author    : ben
 */
#include <stdio.h>
#define SIZE 100
#define MAX 0x7fffffff
typedef struct {
    int Map[SIZE][SIZE];
} City;

int map[SIZE][SIZE];
City city[SIZE];
City tempcity;
int N, M;

void create(int a, int b) {
    int i, j, k, flag;

    for (i = 0; i < M; i++) {
        for (j = 0; j < M; j++) {
            tempcity.Map[i][j] = 0;
            for (k = 0; k < M; k++) {
                tempcity.Map[i][j] += city[a].Map[i][k] * city[b].Map[k][j];
            }
        }
    }

    for (k = 0; k < N; k++) {
        if(k == a || k == b) {
            continue;
        }
        flag = 0;
        for (i = 0; i < M; i++) {
            if(flag) {
                break;
            }
            for (j = 0; j < M; j++) {
                if (city[k].Map[i][j] != tempcity.Map[i][j]) {
                    flag = 1;
                    break;
                }
            }
        }
        if(!flag) {
            map[a][k] = 1;
        }
    }
}

void init() {
    int i, j, k;
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            map[i][j] = MAX;
        }
        map[i][i] = 0;
    }
    for (k = 0; k < N; k++) {
        for (i = 0; i < M; i++) {
            for (j = 0; j < M; j++) {
                scanf("%d", &city[k].Map[i][j]);
            }
        }
    }
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            if (i != j) {
                create(i, j);
            }
        }
    }
}

void Floyd() {
    int i, j, k;
    for (k = 0; k < N; k++) {
        for (i = 0; i < N; i++) {
            if (i == k || map[i][k] == MAX) {
                continue;
            }
            for (j = 0; j < N; j++) {
                if (k == j || i == j || map[k][j] == MAX) {
                    continue;
                }
                if (map[i][k] < map[i][j] - map[k][j]) {
                    map[i][j] = map[i][k] + map[k][j];
                }
            }
        }
    }
}

void query() {
    int Q, i, j;
    scanf("%d", &Q);
    while (Q--) {
        scanf("%d %d", &i, &j);
        if (map[i - 1][j - 1] < MAX) {
            printf("%d\n", map[i - 1][j - 1]);
        } else {
            printf("Sorry\n");
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    while (scanf("%d %d", &N, &M) == 2) {
        if (N + M == 0) {
            break;
        }
        init();
        Floyd();
        query();
    }
    return 0;
}