uva 646 - The Gourmet Club(暴力)

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

;
bool flag;
int g

, ans[8][4];
int c
, can

, v
;

bool init () {
flag = false;
char str
;
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;
}