因为cf上一堆水题，每个单独开一篇博客感觉不太好，就直接放一起好了。
CF1096D Easy Problem
给定字符串，每个位置删除要代价。求最小代价使之不含子序列"hard"。
设f[i][f]表示前i个删到只匹配f位子序列的最小代价。转移看代码吧。O(n)
1 #include <bits/stdc++.h> 2 3 typedef long long LL; 4 const int N = 100010; 5 6 int a[N]; 7 LL f[N][5]; 8 char str[N]; 9 10 int main() { 11 int n; 12 scanf("%d", &n); 13 scanf("%s", str + 1); 14 for(int i = 1; i <= n; i++) scanf("%d", &a[i]); 15 16 int tag = 0; 17 for(int i = 1; i <= n; i++) { 18 if(tag == 0 && str[i] == 'h') tag++; 19 if(tag == 1 && str[i] == 'a') tag++; 20 if(tag == 2 && str[i] == 'r') tag++; 21 if(tag == 3 && str[i] == 'd') tag++; 22 } 23 if(tag != 4) { 24 printf("0 "); 25 return 0; 26 } 27 28 for(int i = 1; i <= n; i++) { 29 f[i][0] = f[i - 1][0] + a[i] * (str[i] == 'h'); 30 f[i][1] = f[i - 1][1] + a[i] * (str[i] == 'a'); 31 f[i][2] = f[i - 1][2] + a[i] * (str[i] == 'r'); 32 f[i][3] = f[i - 1][3] + a[i] * (str[i] == 'd'); 33 if(str[i] == 'h') f[i][1] = std::min(f[i][1], f[i - 1][0]); 34 if(str[i] == 'a') f[i][2] = std::min(f[i][2], f[i - 1][1]); 35 if(str[i] == 'r') f[i][3] = std::min(f[i][3], f[i - 1][2]); 36 } 37 38 LL ans = std::min(std::min(f[n][0], f[n][1]), std::min(f[n][2], f[n][3])); 39 40 printf("%lld ", ans); 41 return 0; 42 }
CF1036C Classy Numbers
问[l, r]之间有多少个数满足非0数码不超过三个。
裸数位DP......注意每次读完要把char数组清空...
1 /** 2 * There is no end though there is a start in space. ---Infinity. 3 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite. 4 * Only the person who was wisdom can read the most foolish one from the history. 5 * The fish that lives in the sea doesn't know the world in the land. 6 * It also ruins and goes if they have wisdom. 7 * It is funnier that man exceeds the speed of light than fish start living in the land. 8 * It can be said that this is an final ultimatum from the god to the people who can fight. 9 * 10 * Steins;Gate 11 */ 12 13 #include <bits/stdc++.h> 14 15 typedef long long LL; 16 const int N = 25; 17 18 LL f[N][5][2]; 19 char str[N], str1[N], str2[N]; 20 21 LL DFS(int k, int cnt, int z) { 22 if(cnt > 3) return 0; 23 if(!k) { 24 return 1; 25 } 26 if(f[k][cnt][z] != -1) { 27 return f[k][cnt][z]; 28 } 29 LL ans = 0; 30 if(cnt < 3) { 31 for(int i = 1; i <= 9; i++) { 32 ans += DFS(k - 1, cnt + 1, 1); 33 } 34 } 35 ans += DFS(k - 1, cnt, z); 36 return f[k][cnt][z] = ans; 37 } 38 39 LL DFS_2(int k, int cnt, int z) { 40 if(cnt > 3) return 0; 41 if(!k) return 1; 42 int lm = str[k] - '0'; 43 //printf("k = %d cnt = %d z = %d lm = %d ", k, cnt, z, lm); 44 LL ans = DFS_2(k - 1, cnt + (lm != 0), z | (lm != 0)); 45 if(lm) { 46 ans += DFS(k - 1, cnt, z); 47 } 48 for(int i = 1; i < lm; i++) { 49 ans += DFS(k - 1, cnt + 1, 1); 50 } 51 return ans; 52 } 53 /* 54 1 55 29000000000091 29000000000099 56 57 */ 58 inline void dec(int &n) { 59 for(int i = 1; i <= n; i++) { 60 if(str[i] != '0') { 61 str[i]--; 62 break; 63 } 64 else str[i] = '9'; 65 } 66 if(str[n] == '0') n--; 67 return; 68 } 69 70 int main() { 71 memset(f, -1, sizeof(f)); 72 LL l, r; 73 int T; 74 scanf("%d", &T); 75 for(int i = 1; i <= T; i++) { 76 scanf("%s%s", str1 + 1, str2 + 1); 77 int n = strlen(str2 + 1); 78 memcpy(str + 1, str2 + 1, sizeof(char) * n); 79 std::reverse(str + 1, str + n + 1); 80 81 /*for(int j = 1; j <= n; j++) { 82 putchar(str[j]); 83 } 84 puts("");*/ 85 86 LL ans = DFS_2(n, 0, 0); 87 //printf("temp ans = %lld ", ans); 88 memset(str2 + 1, 0, n * sizeof(char)); 89 n = strlen(str1 + 1); 90 memcpy(str + 1, str1 + 1, n * sizeof(char)); 91 std::reverse(str + 1, str + n + 1); 92 dec(n); 93 94 /*for(int j = 1; j <= n; j++) { 95 putchar(str[j]); 96 } 97 puts("");*/ 98 99 ans -= DFS_2(n, 0, 0); 100 printf("%lld ", ans); 101 memset(str1 + 1, 0, n * sizeof(char)); 102 } 103 104 return 0; 105 }
CF1132C Painting the Fence
给定n个区间，选出n - 2个区间使它们覆盖的总长度最大。转为删去两个区间。
去重 + 排序。考虑到答案要么相邻要么不相邻，相邻的预处理前后缀覆盖总长之后直接枚举。
预处理出只删每个区间的时候的覆盖减少量为di。当不相邻的时候，考虑到减少总量就是他们两个的di之和。
于是枚举一个区间，拿一个数据结构维护[1, i - 2]之间的最小d值即可。
注意答案不能比总长度还优。复杂度O(nlogn) / O(n)
1 /** 2 * There is no end though there is a start in space. ---Infinity. 3 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite. 4 * Only the person who was wisdom can read the most foolish one from the history. 5 * The fish that lives in the sea doesn't know the world in the land. 6 * It also ruins and goes if they have wisdom. 7 * It is funnier that man exceeds the speed of light than fish start living in the land. 8 * It can be said that this is an final ultimatum from the god to the people who can fight. 9 * 10 * Steins;Gate 11 */ 12 13 #include <bits/stdc++.h> 14 15 const int N = 5010; 16 17 struct Node { 18 int l, r; 19 inline bool operator < (const Node &w) const { 20 if(l != w.l) return l < w.l; 21 return r > w.r; 22 } 23 }node[N], stk[N]; int top; 24 25 int n, m, sl[N], sr[N], del[N], pw[N], ST[N][20]; 26 27 namespace t1 { 28 int d[N]; 29 inline void solve() { 30 for(int i = 1; i <= n; i++) { 31 d[node[i].l]++; 32 d[node[i].r + 1]--; 33 } 34 int ans = 0, now = 0; 35 for(int i = 1; i <= m; i++) { 36 now += d[i]; 37 ans += (now > 0); 38 } 39 printf("%d ", ans); 40 return; 41 } 42 } 43 44 namespace t2 { 45 inline void solve() { 46 int ans = 0; 47 for(int i = 1; i <= n; i++) { 48 ans = std::max(ans, std::min(sl[i - 1] + sr[i + 1], sr[1])); 49 } 50 printf("%d ", ans); 51 return; 52 } 53 } 54 55 inline void prework() { 56 for(int i = 1; i <= n; i++) { /// get sl del 57 sl[i] = sl[i - 1] + node[i].r - std::max(node[i].l, node[i - 1].r + 1) + 1; 58 del[i] = std::min(node[i].r, node[i + 1].l - 1) - std::max(node[i].l, node[i - 1].r + 1) + 1; 59 del[i] = std::max(del[i], 0); 60 } 61 for(int i = n; i >= 1; i--) { /// get sr 62 sr[i] = sr[i + 1] + std::min(node[i].r, node[i + 1].l - 1) - node[i].l + 1; 63 } 64 return; 65 } 66 67 inline void prework2() { 68 for(int i = 2; i <= n; i++) { 69 pw[i] = pw[i >> 1] + 1; 70 } 71 for(int i = 1; i <= n; i++) ST[i][0] = del[i]; 72 for(int j = 1; j <= pw[n]; j++) { 73 for(int i = 1; i + (1 << j) - 1 <= n; i++) { 74 ST[i][j] = std::min(ST[i][j - 1], ST[i + (1 << (j - 1))][j - 1]); 75 } 76 } 77 return; 78 } 79 80 inline int getMin(int l, int r) { 81 if(l > r) return 0x3f3f3f3f; 82 int t = pw[r - l + 1]; 83 return std::min(ST[l][t], ST[r - (1 << t) + 1][t]); 84 } 85 86 int main() { 87 scanf("%d%d", &m, &n); 88 for(int i = 1; i <= n; i++) { 89 scanf("%d%d", &node[i].l, &node[i].r); 90 } 91 92 std::sort(node + 1, node + n + 1); 93 int k = 2; 94 for(int i = 1; i <= n; i++) { 95 if(top && stk[top].r >= node[i].r) { 96 k--; 97 continue; 98 } 99 stk[++top] = node[i]; 100 } 101 n = top; 102 memcpy(node + 1, stk + 1, n * sizeof(Node)); 103 104 if(k <= 0) { 105 t1::solve(); 106 return 0; 107 } 108 node[n + 1].l = node[n + 1].r = 0x3f3f3f3f; 109 prework(); 110 111 if(k == 1) { 112 t2::solve(); 113 return 0; 114 } 115 116 /// solve 117 118 int ans = 0; 119 for(int i = 1; i < n; i++) { /// choose i i + 1 120 ans = std::max(ans, std::min(sl[i - 1] + sr[i + 2], sr[1])); 121 } 122 /// not neighbor 123 124 prework2(); 125 126 for(int i = 1; i <= n; i++) { 127 ans = std::max(ans, sr[1] - del[i] - getMin(1, i - 2)); 128 } 129 printf("%d ", ans); 130 return 0; 131 }
CF1132F Clear the String
给定字符串，一次可以删去连续的一段相同字符，问最少多少次删光。
有一个很直观的想法是f[l][r][k]表示把[l, r]删成k个str[l]的最小代价，发现不好O(1)转移...
然后又考虑f[l][r][k]表示[l, r]右端加上k个str[r]之后删去的最小代价，也不会转移...
尝试n2状态，发现这个k，完全没必要记具体个数，因为无论多少个都是一次删掉。
于是采用CF1025D的套路，fa[l][r]表示把[l, r]删成只剩str[l - 1]的最小代价，fb是str[r + 1]，ans是删光的最小代价。
转移fa的时候考虑str[r]和str[l - 1]是否相同并以此转移。转移ans的时候选出一个位置，左边的fb + 右边的fa转移。
1 #include <bits/stdc++.h> 2 3 const int N = 510; 4 5 int fa[N][N], fb[N][N], ans[N][N]; 6 char str[N]; 7 8 inline void exmin(int &a, const int &b) { 9 if(a > b) a = b; 10 return; 11 } 12 13 int main() { 14 memset(fa, 0x3f, sizeof(fa)); 15 memset(fb, 0x3f, sizeof(fb)); 16 memset(ans, 0x3f, sizeof(ans)); 17 int n; 18 scanf("%d", &n); 19 scanf("%s", str + 1); 20 for(int i = 1; i <= n; i++) { 21 fa[i][i] = (str[i] != str[i - 1]); 22 fb[i][i] = (str[i] != str[i + 1]); 23 ans[i][i] = 1; 24 } 25 for(int len = 2; len <= n; len++) { 26 for(int l = 1; l + len - 1 <= n; l++) { 27 int r = l + len - 1; 28 /// fa[l][r] fb[l][r] ans[l][r] 29 if(str[r] == str[l - 1]) { 30 exmin(fa[l][r], fa[l][r - 1]); 31 } 32 else { 33 for(int k = l; k < r; k++) { 34 exmin(fa[l][r], fa[l][k] + ans[k + 1][r]); 35 } 36 } 37 if(str[l] == str[r + 1]) { 38 exmin(fb[l][r], fb[l + 1][r]); 39 } 40 else { 41 for(int k = l; k < r; k++) { 42 exmin(fb[l][r], ans[l][k] + fb[k + 1][r]); 43 } 44 } 45 for(int k = l + 1; k < r; k++) { 46 exmin(ans[l][r], fb[l][k - 1] + fa[k + 1][r] + 1); 47 } 48 exmin(ans[l][r], fa[l + 1][r] + 1); 49 exmin(ans[l][r], fb[l][r - 1] + 1); 50 exmin(fa[l][r], ans[l][r]); 51 exmin(fb[l][r], ans[l][r]); 52 //printf("[%d %d] l = %d r = %d ans = %d ", l, r, fa[l][r], fb[l][r], ans[l][r]); 53 } 54 } 55 printf("%d ", ans[1][n]); 56 return 0; 57 }
然而本题还有多种简单一点的方法......
就设f[l][r]表示把[l, r]删光的代价，考虑str[l]是如何被删的。显然要么是单独被删，要么是跟某些位置一起删。不妨设一起删的位置中第二个是k。
枚举这个k，于是答案就是f[l + 1][k - 1] + f[k][r]
1 #include <bits/stdc++.h> 2 3 const int N = 510; 4 5 int f[N][N]; 6 char str[N]; 7 8 inline void exmin(int &a, const int &b) { 9 if(a > b) a = b; 10 return; 11 } 12 13 int main() { 14 memset(f, 0x3f, sizeof(f)); 15 int n; 16 scanf("%d", &n); 17 scanf("%s", str + 1); 18 for(int i = 1; i <= n; i++) { 19 f[i][i] = 1; 20 } 21 for(int len = 2; len <= n; len++) { 22 for(int l = 1; l + len - 1 <= n; l++) { 23 int r = l + len - 1; 24 /// fa[l][r] fb[l][r] ans[l][r] 25 exmin(f[l][r], f[l + 1][r] + 1); 26 if(str[l] == str[l + 1]) { 27 exmin(f[l][r], f[l + 1][r]); 28 } 29 for(int k = l + 2; k <= r; k++) { 30 if(str[k] == str[l]) { 31 exmin(f[l][r], f[l + 1][k - 1] + f[k][r]); 32 } 33 } 34 } 35 } 36 printf("%d ", f[1][n]); 37 return 0; 38 }
CF1132D Stressful Training
题意：有n个笔记本，一开始有ai格电，每分钟消耗bi格电。你要买一个每分钟充x格点的插头，使得这些笔记本都能撑过k分钟。问x的最小值。
解：不难想到二分。然后check里每天挑最早没电的充电。如果直接用堆的话是nlog2n的，然而我卡过去了...
实际上我们可以考虑把这些笔记本按照没电天来分类，发现同类笔记本我们不需要知道它们的相对大小，于是直接vector。然后从前往后扫一遍，就能做到线性。
注意二分上界是1e12...
1 #include <bits/stdc++.h> 2 3 typedef long long LL; 4 const int N = 200010; 5 6 inline char gc() { 7 static char buf[N], *p1(buf), *p2(buf); 8 if(p1 == p2) p2 = (p1 = buf) + fread(buf, 1, N, stdin); 9 return p1 == p2 ? EOF : *p1++; 10 } 11 12 template <class T> inline void read(T &x) { 13 x = 0; 14 register char c(gc()); 15 bool f(false); 16 while(c < '0' || c > '9') { 17 if(c == '-') f = true; 18 c = gc(); 19 } 20 while(c >= '0' && c <= '9') { 21 x = x * 10 + c - 48; 22 c = gc(); 23 } 24 if(f) x = (~x) + 1; 25 return; 26 } 27 28 struct Node { 29 LL now, del; 30 double x; 31 Node(LL A = 0, LL B = 0) { 32 now = A; 33 del = B; 34 x = (double)A / B; 35 } 36 inline bool operator < (const Node &w) const { 37 return x > w.x; 38 } 39 }node[N]; 40 41 int n, k; 42 LL a[N], b[N]; 43 std::priority_queue<Node> Q; 44 45 inline bool check(LL x) { 46 while(Q.size()) Q.pop(); 47 for(register int i = 1; i <= n; i++) { 48 Q.push(node[i]); 49 } 50 for(register int i = 0; i < k; i++) { 51 Node temp = Q.top(); 52 Q.pop(); 53 if(temp.now - temp.del * i < 0) { 54 return false; 55 } 56 temp = Node(temp.now + x, temp.del); 57 Q.push(temp); 58 } 59 Node temp = Q.top(); 60 if(temp.now - temp.del * k < 0) { 61 return false; 62 } 63 return true; 64 } 65 66 int main() { 67 68 read(n); read(k); 69 k--; 70 for(register int i = 1; i <= n; i++) { 71 read(a[i]); 72 } 73 for(register int i = 1; i <= n; i++) { 74 read(b[i]); 75 } 76 for(register int i = 1; i <= n; i++) { 77 if(b[i] * k <= a[i]) { 78 std::swap(a[i], a[n]); 79 std::swap(b[i], b[n]); 80 n--; 81 i--; 82 } 83 } 84 85 if(!n) { 86 puts("0"); 87 return 0; 88 } 89 90 for(int i = 1; i <= n; i++) { 91 node[i] = Node(a[i], b[i]); 92 } 93 std::sort(node + 1, node + n + 1); 94 95 LL l = 1, r = 2e12; 96 while(l < r) { 97 LL mid = (l + r) >> 1; 98 if(check(mid)) { 99 r = mid; 100 } 101 else l = mid + 1; 102 } 103 if(r == 2e12) puts("-1"); 104 else { 105 printf("%lld ", r); 106 } 107 return 0; 108 }