E - Field of Wonders 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest
2017-10-22 18:57
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题意有点难理解
最后的意思就是 能不能找一个字符 使得原串中隐藏位置的字符显现出来,问这样的字符可能有几个
首先,给定的 m 个串可能是不合法的,排除掉;
而且 m 个中至少有一个是跟远原来串一样的,但是 m 个串中有的有 有的没有的字符 不确定是不是一定存在,,所以要找 m 个串中都存在的字符才能实现题意要求
所以 最后也就是求这样的 字符的个数
#include<bits/stdc++.h>
using namespace std;
const int maxn = 50 + 7;
int n, m, cnt;
char p[maxn], s[1001][maxn];
int a[maxn];
bool vis[1001];
set<char> st1;
void init() {
scanf("%d", &n);
scanf("%s", p);
cnt = 0;
for(int i = 0; i < n; ++i) {
if(p[i] == '*') a[cnt++] = i;
else st1.insert(p[i]);
}
scanf("%d", &m);
memset(vis, false, sizeof vis);
for(int i = 0; i < m; ++i) {
scanf("%s", s[i]);
int f = 0;
for(int j = 0; j < n; ++j) {
if(p[j] == '*') {
if(st1.count(s[i][j])) { f = 1; break; }
}
else if(p[j] != s[i][j]) { f = 1; break; }
}
if(f) vis[i] = true;
}
}
bool is_ok(char c) {
int t = cnt;
int tt = min(1, cnt-1);
for(int i = 0; i < m; ++i) {
if(vis[i]) continue;
t = cnt;
for(int j = 0; j < cnt; ++j) {
if(s[i][a[j]] == c) {t--; break; }
}
if(t == cnt) return false;
}
return true;
}
void solve() {
int ans = 0;
for(int i = 0; i < 26; ++i) {
char c = 'a' + i;
if(st1.count(c)) continue;
else {
if(is_ok(c)) ans++;
}
}
cout << ans << endl;
}
int main() {
init();
solve();
return 0;
}
最后的意思就是 能不能找一个字符 使得原串中隐藏位置的字符显现出来,问这样的字符可能有几个
首先,给定的 m 个串可能是不合法的,排除掉;
而且 m 个中至少有一个是跟远原来串一样的,但是 m 个串中有的有 有的没有的字符 不确定是不是一定存在,,所以要找 m 个串中都存在的字符才能实现题意要求
所以 最后也就是求这样的 字符的个数
#include<bits/stdc++.h>
using namespace std;
const int maxn = 50 + 7;
int n, m, cnt;
char p[maxn], s[1001][maxn];
int a[maxn];
bool vis[1001];
set<char> st1;
void init() {
scanf("%d", &n);
scanf("%s", p);
cnt = 0;
for(int i = 0; i < n; ++i) {
if(p[i] == '*') a[cnt++] = i;
else st1.insert(p[i]);
}
scanf("%d", &m);
memset(vis, false, sizeof vis);
for(int i = 0; i < m; ++i) {
scanf("%s", s[i]);
int f = 0;
for(int j = 0; j < n; ++j) {
if(p[j] == '*') {
if(st1.count(s[i][j])) { f = 1; break; }
}
else if(p[j] != s[i][j]) { f = 1; break; }
}
if(f) vis[i] = true;
}
}
bool is_ok(char c) {
int t = cnt;
int tt = min(1, cnt-1);
for(int i = 0; i < m; ++i) {
if(vis[i]) continue;
t = cnt;
for(int j = 0; j < cnt; ++j) {
if(s[i][a[j]] == c) {t--; break; }
}
if(t == cnt) return false;
}
return true;
}
void solve() {
int ans = 0;
for(int i = 0; i < 26; ++i) {
char c = 'a' + i;
if(st1.count(c)) continue;
else {
if(is_ok(c)) ans++;
}
}
cout << ans << endl;
}
int main() {
init();
solve();
return 0;
}
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