Educational Codeforces Round 44 (Rated for Div. 2)
题解：看着像优化问题，实际上因为奇数的位置或者偶数的位置一定会被填满，所以暴力的对每个元素寻找它的适合位置。
感受：比赛的时候想多了，如果棋子的个数较少的话感觉会有点麻烦。。。
#pragma warning(disable:4996) #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define lson root<<1 #define rson root<<1|1 #define mem(arr,in) memset(arr,in,sizeof(arr)) using namespace std; const int maxn = 105; int n; int a[maxn], use[maxn]; int ca_1() { mem(use, 0); for (int i = 1; i <= n; i++) if (a[i] % 2) use[a[i]] = 1; int cnt = 0; for (int i = 1; i <= n; i++) if (a[i] % 2 == 0) { for (int j = 1; j <= 2*n; j += 2) if (!use[j]) { cnt += abs(j - a[i]); use[j] = 1; break; } } return cnt; }
int ca_2() { mem(use, 0); for (int i = 1; i <= n; i++) if (a[i] % 2 == 0) use[a[i]] = 1; int cnt = 0; for (int i = 1; i <= n; i++) if (a[i] % 2) { for (int j = 2; j <= 2*n; j += 2) if (!use[j]) { cnt += abs(j - a[i]); use[j] = 1; break; } } return cnt; } int main() { while (cin >> n) { n = n / 2; for (int i = 1; i <= n; i++) { cin >> a[i]; } sort(a + 1, a + n + 1); int ans = min(ca_1(), ca_2()); cout << ans << endl; } return 0; }
题解：很直白的一道题，判断删除一个开关会影响那盏灯的状态。
感受：这题出的很晚很无奈，，，，
#pragma warning(disable:4996) #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define lson root<<1 #define rson root<<1|1 #define mem(arr,in) memset(arr,in,sizeof(arr)) using namespace std; const int maxn = 2005; int n, m; int mp[maxn][maxn], d[maxn]; char p[maxn][maxn]; int main() { while(scanf("%d%d",&n,&m)!=EOF){ for (int i = 0; i < n; i++) scanf("%s", p[i]); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (p[i][j] == '1') mp[i][j] = 1; else mp[i][j] = 0; } } mem(d, 0); for (int j = 0; j < m; j++) { int cnt = 0; for (int i = 0; i < n; i++) { if (mp[i][j] == 1) cnt++; } d[j] = cnt; } bool flag; for (int i = 0; i < n; i++) { flag = true; for (int j = 0; j < m; j++) { if (mp[i][j] == 1) { if (d[j] <= 1) flag = false; } else { if (d[j] == 0) flag = false; } } if (flag) break; } if (flag) printf("YES\n"); else printf("NO\n"); } return 0; }
题解：我的想法是先排序得到最优的情况，判断是否满足，如果从位置pos就不满足的话，就从pos往前找后面剩下的桶（need）的价值，同时在从前往后找每个桶的价值。先计算前面几个桶的价值（n-need)，在从后往前找剩下的几个桶的价值（need），确保一定有n个桶。具体看代码。。。
感受：翻车最严重的一道题，，，同样的贪心思想，代码却写挂了，死活还找不到错误。。。注释掉的是我的代码，如果有朋友知道哪错了，麻烦告知一声，终于找到错误了，爽！！！！！！
#pragma warning(disable:4996) #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define lson root<<1 #define rson root<<1|1 #define mem(arr,in) memset(arr,in,sizeof(arr)) using namespace std; const int maxn = 100005; int n, k, l, m; int a[maxn]; bool use[maxn]; int main() { while (scanf("%d %d %d", &n, &k, &l) != EOF) { m = n * k; for (int i = 1; i <= m; i++) scanf("%d", &a[i]); sort(a + 1, a + m + 1); if (a[n] - a[1] > l) printf("0\n"); else { ll ans = 0; int pos = m; for (int i = n; i <= m; i++) if (a[i] - a[1] > l) { pos = i - 1; break; } mem(use, 0); int cnt = 0; for (int i = 1; i <= pos; i += k) { ans += a[i]; cnt++; use[i] = 1; } if (cnt == n) printf("%I64d\n", ans); else { for (int i = pos; i >= 1; i--) if (!use[i]) { ans += a[i]; cnt++; if (cnt == n) break; } printf("%I64d\n", ans); } } /* int pos = -1; for (int i = 1; i <= m; i +=k) if (a[i] - a[1] > l) { pos = i; break; } ll ans = 0; if (pos == -1) { for (int i = 1; i <= m; i += k) ans += a[i]; printf("%I64d\n", ans); } else { int need = n - (pos - 1) / k; int cnt = 0; int x = -1; for (int i = pos; i >= 1; i--) if (a[i] - a[1] <= l) { cnt++; ans += a[i]; if (cnt == need) { x = i; break; } } for (int i = 1; i < x; i += k) { ans += a[i]; cnt++; } 这样的顺序不能保证凑成n个桶，实际上当a[n]-a[1]<=l，一定能凑成n个桶。！！！！！ if (cnt != n) printf("0\n"); else printf("%I64d\n", ans); } */ } return 0; }
感受：题意不太好懂，这个题的做法很有意思，找一种构造直接满足题意的很难，但是找一种构造使某个范围内的n都满足题意的却很简单，严格的证明要去看cf提供的标解~~~~，这次受教了。
题解：因为这个盒子内任意两个元素之差要小于d，故先排序。假设存在答案，那么这个答案可以看作是这个序列被分作了几个区间。dp[ i ] = 1 表示第 i 个元素能够成为某个区间的第一个元素。
dp[ i + 1 ] = 1 only if some j , i - j + 1 >= k, and dp[ j ] = 1, and a[ i ] - a[ j ] <= d。so if dp[ n + 1 ] = 1，have answer。
1 #pragma warning(disable:4996) 2 #include<string> 3 #include<map> 4 #include<cstdio> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 #define ll long long 9 #define lson root<<1 10 #define rson root<<1|1 11 #define mem(arr,in) memset(arr,in,sizeof(arr)) 12 using namespace std; 13 typedef pair<int, int> P; 14 15 const int maxn = 500005; 16 17 int n, k, d; 18 int a[maxn], dp[maxn], co[maxn]; 19 20 int Lowbit(int x) { return x & -x; } 21 22 void add(int pos) { 23 for (int i = pos; i <= n; i += Lowbit(i)) co[i] ++; 24 return; 25 } 26 27 int sum(int pos) { 28 int res = 0; 29 for (int i = pos; i >= 1; i -= Lowbit(i)) res += co[i]; 30 return res; 31 } 32 33 int get(int l, int r) { 34 if (l > r) return 0; 35 return sum(r) - sum(l - 1); 36 } 37 38 int main() 39 { 40 while (scanf("%d %d %d", &n, &k, &d) != EOF) { 41 mem(co, 0); 42 mem(dp, 0); 43 dp[1] = 1; 44 add(1); 45 46 for (int i = 1; i <= n; i++) scanf("%d", a + i); 47 sort(a + 1, a + n + 1); 48 49 int l = 1; 50 for (int i = 1; i <= n; i++) { 51 while (l < i && a[i] - a[l] > d) l++; 52 dp[i + 1] = (get(l, i - k + 1) >= 1); 53 if (dp[i + 1]) add(i + 1); 54 } 55 56 //for (int i = 1; i <= n; i++) printf("%d ", dp[i]); 57 //cout << endl; 58 59 puts(dp[n + 1] ? "YES" : "NO"); 60 } 61 return 0; 62 }
