数据有点水,加个卡时就可以过。
代码加注释送上:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ctime>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
bool tong[5][100]; //tong[花色][点数]
struct zzz {
int c, k;
}a[100]; //c:花色,k:点数
bool flag;
int n, ans = 100, ans2 = 100, kti;
bool judge() { //判断是否为顺子
int num = 0; bool flag2 = 1;
for(int i = 1; i <= 4; ++i) {
int minn = 100, maxn = 0;
for(int j = 1; j <= 13; ++j) {
if(tong[i][j] && minn == 100) minn = j;
if(tong[i][j]) maxn = max(maxn, j);
}
for(int j = minn; j <= maxn; ++j)
if(!tong[i][j]) flag2 = 0, ++num;
}
ans2 = min(ans2, num);
return flag2;
}
void dfs(int pos, int tot) {
if(tot >= ans) return ;
if(judge()) {
ans = min(ans, tot);
return ;
}
if(pos > n) return ;
//=====卡时
if((clock() - kti) * 1000 >= 980 * CLOCKS_PER_SEC){
if(ans <= 52) printf("Yes\n%d", ans);
else printf("No\n%d", ans2);
exit(0);
}
//=====
for(int i = 1; i <= 4; ++i) {
if(tong[i][a[pos].k] && i != a[pos].c) continue;
tong[a[pos].c][a[pos].k] = 0;
tong[i][a[pos].k] = 1;
if(i != a[pos].c) dfs(pos+1, tot+1);
else dfs(pos+1, tot);
tong[i][a[pos].k] = 0, tong[a[pos].c][a[pos].k] = 1;
}
}
int main() {
kti = clock(), n = read();
for(int i = 1; i <= n; ++i) {
a[i].c = read();
char c[2]; cin >> c;
if(c[0] == 'A') a[i].k = 1;
else if(c[0] == 'J') a[i].k = 11;
else if(c[0] == 'Q') a[i].k = 12;
else if(c[0] == 'K') a[i].k = 13;
else if(c[0] == '1') a[i].k = 10;
else a[i].k = c[0] - '0';
}
for(int i = 1; i <= n; ++i) tong[a[i].c][a[i].k] = 1;
dfs(1, 0);
if(ans <= 52) printf("Yes\n%d", ans);
else printf("No\n%d", ans2);
return 0;
}
标签:luoguP4907,LL,卡时,else,ans,return,problem,include
From: https://www.cnblogs.com/mcr130102/p/18386933