局域网及网站建设内容,网站建设初学,做网站 知乎,网站建站代码ps#xff1a;比赛的时候想到了做法#xff0c;k次排序#xff0c;然后每次消去能消的。。。然而这种做法是错误的#xff0c;神奇的是测试案例中排在奇数的案例会WA#xff0c;排在偶数的案例都过了#xff0c;被注释的代码会T. #includebits/stdc.h
#define UL… ps比赛的时候想到了做法k次排序然后每次消去能消的。。。然而这种做法是错误的神奇的是测试案例中排在奇数的案例会WA排在偶数的案例都过了被注释的代码会T. #includebits/stdc.h
#define ULL unsigned long long
#define LL long long
#define P pairint, int
#define pb push_back
#define mp make_pair
#define pp pop_back
#define lson root 1
#define INF32 (int)2e9 7
#define rson root 1 | 1
#define INF64 (unsigned long long)1e18
#define sc(x) scanf(%d, x)
#define pr(x) printf(%d\n, x)
#define mem(arry, in) memset(arry, in, sizeof(arry))
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;namespace fastIO {#define BUF_SIZE 100000bool IOerror 0;inline char nc() {static char buf[BUF_SIZE], *p1 buf BUF_SIZE, *pend buf BUF_SIZE;if(p1 pend) {p1 buf;pend buf fread(buf, 1, BUF_SIZE, stdin);if(pend p1) {IOerror 1;return -1;}}return *p1;}inline bool blank(char ch) {return ch || ch \n || ch \r || ch \t;}inline void read(int x) {char ch;while(blank(ch nc()));if(IOerror) return;for(x ch - 0; (ch nc()) 0 ch 9; x x * 10 ch - 0);}#undef BUF_SIZE
};
using namespace fastIO;inline void upd(int x, int y) { x y (x y); }const int N 100005;P a[15][N];
int T, n, k, v[10], b[N][15], point[15], cnt[N], c[N][15];
bool use[N];int main()
{//freopen(D:\\1.in, r, stdin);//freopen(D:\\1.txt, w, stdout);read(T);while(T--) {read(n), read(k);for (int i 1; i k; i) read(v[i]);for (int i 1; i n; i) {for (int j 1; j k; j) {read(a[j][i].first);a[j][i].second i;//c[i][j] a[j][i].first;}for (int j 1; j k; j) read(b[i][j]);}mem(cnt, 0);mem(point, 0);//mem(use, 0);int ans 0;for (int i 1; i k; i) sort(a[i] 1, a[i] 1 n);/* // 写的代码真的很搓也许换种优秀的写法就过了sort(a[1] 1, a[1] n 1);while(1) {int oldans ans;for (int j 1; j n; j) if (!use[a[1][j].second]) {int id a[1][j].second;if (c[id][1] v[1]) break;int tot 0;for (int q 1; q k; q) if (c[id][q] v[q]) {tot;}if (tot k) {for (int q 1; q k; q) v[q] b[id][q];use[id] 1;ans;}}if (oldans ans) break;}*/while(1) {int oldans ans;for (int i 1; i k; i) {while(point[i] n a[i][point[i] 1].first v[i]) {int x a[i][point[i]].second;cnt[x];if (cnt[x] k) {ans;for (int j 1; j k; j) v[j] b[x][j];}}}if (oldans ans) break;}pr(ans);for (int i 1; i k; i) printf(%d , v[i]);pr(v[k]);}return 0;
} 转载于:https://www.cnblogs.com/zgglj-com/p/9474734.html
