题目链接：uva 646 - The Gourmet Club

题目大意：有16个人参加聚会，聚会一共5天，每天有4桌，每桌4个人，一起吃饭的4个人会互相认识。现在要安排座位使得16个任意两个人都互相认识。给出前三天的安排，求后两天的安排。

解题思路：任意两个人之间肯定只能同桌一次。所以根据这个条件，只要枚举出第4天的第1桌的情况，就可推导出所有的，或者是矛盾。

在Poj和Zoj上都过了，uva上过不了，求大神指教。

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
const int N = 20;

char s[N][N];
bool flag;
int g[N][N], ans[8][4];
int c[N], can[N][N], v[N];

bool init () {
    flag = false;
    char str[N];
    memset(g, 0, sizeof(g));
    memset(c, 0, sizeof(c));

    for (int i = 0; i < 3; i++) {
        memset(v, 0, sizeof(v));
        for (int j = 0; j < 4; j++) {

            if (scanf("%s", str) == 1) {

                for (int t = 0; t < 4; t++) {
                    if (v[str[t]-‘A‘]) flag = true;
                    v[str[t]-‘A‘] = 1;

                    for (int k = 0; k < 4; k++) {
                        if (t == k) continue;
                        g[str[t]-‘A‘][str[k]-‘A‘]++;
                    }
                    g[str[t]-‘A‘][str[t]-‘A‘] = 1;
                }
            } else 
                return false;
            memcpy(s[i*4+j], str, sizeof(str));
        }
    }


    for (int i = 0; i < 16; i++) {
        for (int j = 0; j < 16; j++) {
            if (g[i][j] == 0) {
                can[i][c[i]++] = j;
            }
        }
    }
    memset(v, 0, sizeof(v));
    return true;
}

int getX (int* x) {
    for (int i = 0; i < 4; i++)
        if (v[x[i]] == 1) return x[i];
    return -1;
}

void putAns() {
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            if (j) printf("    ");
            printf("%s", s[i*4+j]);
        }
        printf("\n");
    }
    for (int k = 0; k < 2; k++) {
        for (int i = k; i < 8; i += 2) {
            if (i/2) printf("    ");
            for (int j = 0; j < 4; j++)
                printf("%c", ‘A‘ + ans[i][j]);
        }
        printf("\n");
    }
}

bool judge (int x, int* y) {
    for (int i = 0; i < 4; i++)
        if (y[i] == x) return false;
    return true;
}

bool DFS (int d) {

    if (d == 8) {
        putAns();
        return true;
    }

    if (d == 0) {
        do {
            ans[d][0] = 0;
            memcpy (ans[d]+1, can[0], sizeof(int)*3);

            for (int i = 0; i < 4; i++) v[ans[d][i]]++;

            if (DFS(d+1)) return true;

            for (int i = 0; i < 4; i++) v[ans[d][i]]--;

        } while (next_permutation (can[0], can[0] + c[0]));
    } else {
        int x = getX(ans[d-1]);
        if (x == -1) return false;

        int cnt = 1;
        ans[d][0] = x;
        for (int i = 0; i < c[x]; i++) {
            int& u = can[x][i];
            if (v[u] < 2 && judge (u, ans[d-1])) {
                ans[d][cnt++] = can[x][i];
            }
        }

        if (cnt != 4) return false;

        for (int i = 0; i < 4; i++) v[ans[d][i]]++;

        if (DFS(d+1)) return true;

        for (int i = 0; i < 4; i++) v[ans[d][i]]--;
    }

    return false;
}

int main () {
    while (init ()) {

        if (flag || !DFS (0))
            printf("It is not possible to complete this schedule.\n");
        printf("\n");
    }
    return 0;
}