标签：unsigned 密码 type 全局 print scan uva 递归 [1]
题意：给两个6行5列的字母矩阵，找出满足如下条件的“密码”：密码中的每个字母在两个矩阵的对应列中均出现。给定k（1<=k<=7777），你的任务是找出字典序第k小的密码。如果不存在，输出NO。
分析：因为k<=7777，直接按字典序从小到大的顺序递归一个一个的枚举。
注意：定义在dfs里的vis不能放在全局，否则会导致值的混用。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 30000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char pic[2][6][5];
char ans[6];
int k;
int cnt;
bool dfs(int cur){
if(cur == 5){
if(++cnt == k){
ans[cur] = ‘\0‘;
return true;
}
return false;
}
int vis[2][26];
memset(vis, 0, sizeof vis);
for(int i = 0; i < 2; ++i){
for(int j = 0; j < 6; ++j){
vis[i][pic[i][j][cur] - ‘A‘] = 1;
}
}
for(int i = 0; i < 26; ++i){
if(vis[0][i] && vis[1][i]){
ans[cur] = ‘A‘ + i;
if(dfs(cur + 1)) return true;
}
}
return false;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
cnt = 0;
scanf("%d", &k);
for(int i = 0; i < 2; ++i){
for(int j = 0; j < 6; ++j){
scanf("%s", pic[i][j]);
}
}
if(!dfs(0)){
printf("NO\n");
}
else{
printf("%s\n", ans);
}
}
return 0;
}
标签：unsigned 密码 type 全局 print scan uva 递归 [1]
原文地址：http://www.cnblogs.com/tyty-Somnuspoppy/p/6388398.html
