这一道题的正解是倒推（然后有难度吗？）,直接用results[i]表示第i位在某次操作完成后的位置（第几次呢，就看外重循环吧）接着可以分成三种情况
- 当results[j] <= insi时，对位置没有影响
- 当results[j] > insi && results[j] <= insi + endi - begini时，正好被copy走了，所以位置是begini + resultsi - insi
- 以上两种都不是，那就是被copy插进来的部分导致往后移了endi - begini位，所以减去就好了
Code
#include<iostream> #include<cstdio> #include<cctype> #include<cstring> #include<cstdlib> #include<cmath> #include<fstream> #include<sstream> #include<algorithm> #include<map> #include<set> #include<queue> #include<vector> #include<stack> using namespace std; typedef bool boolean; #define INF 0xfffffff #define smin(a, b) a = min(a, b) #define smax(a, b) a = max(a, b) template<typename T> inline void readInteger(T& u){ char x; int aFlag = 1; while(!isdigit((x = getchar())) && x != '-'); if(x == '-'){ x = getchar(); aFlag = -1; } for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); ungetc(x, stdin); u *= aFlag; } inline void getLine(char* str){ int i = 0; while((str[i] = getchar()) != ' ' && ~str[i]) i++; str[i] = 0; } int k, m; int n; char s[200005]; int *from; int *end; int *ins; inline void init(){ readInteger(k); readInteger(m); getchar(); getLine(s + 1); readInteger(n); from = new int[(const int)(n + 1)]; end = new int[(const int)(n + 1)]; ins = new int[(const int)(n + 1)]; for(int i = 1; i <= n; i++){ readInteger(from[i]); readInteger(end[i]); readInteger(ins[i]); } } int *results; inline void solve(){ results = new int[(const int)(k + 1)]; for(int i = 1; i <= k; i++) results[i] = i; for(int i = n; i >= 1; i--){ for(int j = 1; j <= k; j++){ int& x = results[j]; if(x <= ins[i]) continue; //没有影响 else if(x <= ins[i] + (end[i] - from[i])){ x = from[i] + x - ins[i]; }else{ x -= end[i] - from[i]; } } } for(int i = 1; i <= k; i++) putchar(s[results[i]]); } int main(){ freopen("A.in", "r", stdin); freopen("A.out", "w", stdout); init(); solve(); return 0; }
这道题是一道水题，首先可以这么想，对于每个si的前4i - 1都是可以通过计算得到了，那我们弄三个前缀和，统计J、O、I分别出现的次数，接着递归调用去计算就好了。如果到了最后一位，不用管，什么都可以。最后找一个最大值，然后就可以ac了。
Code
#include<iostream> #include<cstdio> #include<cctype> #include<cstring> #include<cstdlib> #include<cmath> #include<fstream> #include<sstream> #include<algorithm> #include<map> #include<set> #include<queue> #include<vector> #include<stack> using namespace std; typedef bool boolean; #define INF 0xfffffff #define smin(a, b) a = min(a, b) #define smax(a, b) a = max(a, b) template<typename T> inline void readInteger(T& u){ char x; int aFlag = 1; while(!isdigit((x = getchar())) && x != '-'); if(x == '-'){ x = getchar(); aFlag = -1; } for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); ungetc(x, stdin); u *= aFlag; } int power[11]; inline void getPower(){ power[0] = 1; for(int i = 1; i <= 10; i++) power[i] = power[i - 1] * 4; } const char ch[3] = {'J', 'O', 'I'}; int n; char* str; inline void init(){ readInteger(n); str = new char[power[n] + 1]; getchar(); for(int i = 1; i <= power[n]; i++) str[i] = getchar(); } int *sj; int *so; int *si; inline void presum(){ sj[0] = so[0] = si[0] = 0; for(int i = 1; i <= power[n]; i++){ sj[i] = sj[i - 1], so[i] = so[i - 1], si[i] = si[i - 1]; if(str[i] == 'J') sj[i]++; else if(str[i] == 'O') so[i]++; else if(str[i] == 'I') si[i]++; } for(int i = 1; i <= power[n]; i++){ int j = power[n] + i; sj[j] = sj[j - 1], so[j] = so[j - 1], si[j] = si[j - 1]; if(str[i] == 'J') sj[j]++; else if(str[i] == 'O') so[j]++; else if(str[i] == 'I') si[j]++; } } int *replaces; inline void solve2(int n, int b){ if(n == 0) return; int seg = power[n - 1]; for(int i = 1; i <= power[::n]; i++){ replaces[i] += (seg - (sj[i + b + seg - 1] - sj[i + b - 1])) + (seg - (so[i + b +seg * 2 - 1] - so[i + b + seg - 1])) + (seg - (si[i + b + seg * 3 - 1] - si[i + b + seg * 2 - 1])); } solve2(n - 1, b + seg * 3); } int main(){ freopen("B.in", "r", stdin); freopen("B.out", "w", stdout); getPower(); init(); sj = new int[(const int)(2 * power[n] + 1)]; so = new int[(const int)(2 * power[n] + 1)]; si = new int[(const int)(2 * power[n] + 1)]; presum(); replaces = new int[(const int)(power[n] + 1)]; memset(replaces, 0, sizeof(int) * (power[n] + 1)); solve2(n, 0); int minv = power[n]; for(int i = 1; i <= power[n]; i++) smin(minv, replaces[i]); printf("%d", minv); return 0; }
这道题算是今天最有价值的一道题，废话不多说，讲解法
如果不知道如何dp的可以看看这个帖子合唱队形,dp的话就不用管那个不拔，但是却不产生作用的，只管像这样做，只把i和j中比j高的拔掉就行了。于是得到了dp方程:
中间那一串求和其实可以先预处理出来，但是时间复杂度是O(n2)，还是过不去，这时可以考虑用数据结构维护。首先要按照高度来查询对不对，又要区间更新(拔掉这一棵)，还要单点修改，然后还要区间求最值，是不是可以想到线段树这个好东西。但是内存上过不去对不对？如果按照n的大小来建树内存的开销就很小了。那么可以吧h进行离散化。每次查询就查询1 ~ hi的最值。接着将高度小于等于它的部分全部减去ci。然后单点更新hi的最值。
Notice:
- 第二次dp的时候不要重建树,不然会TLE
- lazy标记和数据都要使用long long
- 注意延时更新的地方不要手抖把+=写成=了
Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cctype> 4 #include<cstring> 5 #include<cstdlib> 6 #include<cmath> 7 #include<fstream> 8 #include<sstream> 9 #include<algorithm> 10 #include<map> 11 #include<set> 12 #include<queue> 13 #include<vector> 14 #include<stack> 15 using namespace std; 16 typedef bool boolean; 17 #define INF 0xfffffff 18 #define smin(a, b) a = min(a, b) 19 #define smax(a, b) a = max(a, b) 20 template<typename T> 21 inline void readInteger(T& u){ 22 char x; 23 int aFlag = 1; 24 while(!isdigit((x = getchar())) && x != '-'); 25 if(x == '-'){ 26 x = getchar(); 27 aFlag = -1; 28 } 29 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 30 ungetc(x, stdin); 31 u *= aFlag; 32 } 33 34 template<typename T> 35 inline void putInteger(T u){ 36 if(u == '0'){ 37 putchar('0'); 38 return; 39 } 40 if(u < 0){ 41 putchar('-'); 42 u *= -1; 43 } 44 stack<char> s; 45 while(u != 0) s.push(u % 10 + '0'), u /= 10; 46 while(!s.empty()) putchar(s.top()), s.pop(); 47 } 48 49 const long long inf = (1ll << 60); 50 51 template<typename T> 52 class TreeNode{ 53 public: 54 int from; 55 int end; 56 TreeNode* left; 57 TreeNode* right; 58 T data; 59 T lazy; 60 TreeNode(int from, int end, T data):from(from), end(end), left(NULL), right(NULL), data(data), lazy(0){ } 61 }; 62 63 template<typename T> 64 class SegTree{ 65 public: 66 TreeNode<T>* root; 67 SegTree():root(NULL){} 68 SegTree(int size){ 69 build(root, 1, size); 70 } 71 inline void pushUp(TreeNode<T>* node){ 72 node->data = max(node->left->data, node->right->data); 73 } 74 inline void pushDown(TreeNode<T>* node){ 75 if(node->left != NULL){ 76 node->left->lazy += node->lazy; 77 node->left->data -= node->lazy; 78 node->right->lazy += node->lazy; 79 node->right->data -= node->lazy; 80 } 81 node->lazy = 0; 82 } 83 void build(TreeNode<T>*& node, int from, int end){ 84 if(from == end){ 85 node = new TreeNode<T>(from, end, 0); 86 return; 87 } 88 node = new TreeNode<T>(from, end, 0); 89 int mid = (from + end) >> 1; 90 build(node->left, from, mid); 91 build(node->right, mid + 1, end); 92 } 93 void update(TreeNode<T>* node, int from, int end, long long val){ 94 if(from == node->from && end == node->end){ 95 node->lazy += val; 96 node->data -= val; 97 return; 98 } 99 if(node->lazy != 0) pushDown(node); 100 int mid = (node->from + node->end) >> 1; 101 if(from > mid) update(node->right, from, end, val); 102 else if(end <= mid) update(node->left, from, end, val); 103 else{ 104 update(node->left, from, mid, val); 105 update(node->right, mid + 1, end, val); 106 } 107 } 108 void update(TreeNode<T>* node, int d, long long val){ 109 if(node->from == d && node->end == d){ 110 smax(node->data, val); 111 return; 112 } 113 if(node->lazy != 0) pushDown(node); 114 int mid = (node->from + node->end) >> 1; 115 if(d > mid) update(node->right, d, val); 116 else update(node->left, d, val); 117 pushUp(node); 118 } 119 long long query(TreeNode<T>* node, int from, int end){ 120 if(from == node->from && end == node->end){ 121 return node->data; 122 } 123 if(node->lazy != 0) pushDown(node); 124 int mid = (node->from + node->end) >> 1; 125 if(from > mid) return query(node->right, from, end); 126 else if(end <= mid) return query(node->left, from, end); 127 else{ 128 return max(query(node->left, from, mid), query(node->right, mid + 1, end)); 129 } 130 } 131 void clean(TreeNode<T>* node){ 132 if(node == NULL) return; 133 clean(node->left); 134 clean(node->right); 135 node->data = 0; 136 node->lazy = 0; 137 } 138 }; 139 140 int n; 141 int *h, *p, *c; 142 143 inline void init(){ 144 readInteger(n); 145 h = new int[(const int)(n + 1)]; 146 p = new int[(const int)(n + 1)]; 147 c = new int[(const int)(n + 1)]; 148 for(int i = 1; i <= n; i++){ 149 readInteger(h[i]); 150 readInteger(p[i]); 151 readInteger(c[i]); 152 } 153 } 154 155 typedef class Data{ 156 public: 157 int height; 158 int index; 159 Data(const int height = 0, const int index = 0):height(height), index(index){ } 160 boolean operator <(Data another) const { 161 return this->height < another.height; 162 } 163 }Data; 164 165 int ch; 166 Data* data; 167 inline void discretization(){ 168 data = new Data[(const int)(n + 1)]; 169 data[0] = Data(0, 0); 170 for(int i = 1; i <= n; i++) 171 data[i] = Data(h[i], i); 172 sort(data + 1, data + n + 1); 173 for(int i = 1; i <= n; i++){ 174 h[data[i].index] = (data[i].height == data[i - 1].height) ? (ch) : (++ch); 175 } 176 } 177 178 long long *lf, *rf; 179 SegTree<long long> st; 180 inline void solve(){ 181 lf = new long long[(const int)(n + 1)]; 182 rf = new long long[(const int)(n + 2)]; 183 lf[0] = rf[n + 1] = 0; 184 st = SegTree<long long>(ch); 185 for(int i = 1; i <= n; i++){ 186 lf[i] = st.query(st.root, 1, h[i]) + p[i]; 187 st.update(st.root, 1, h[i], c[i]); 188 st.update(st.root, h[i], lf[i]); 189 } 190 st.clean(st.root); 191 for(int i = n; i >= 1; i--){ 192 rf[i] = st.query(st.root, 1, h[i]) + p[i]; 193 st.update(st.root, 1, h[i], c[i]); 194 st.update(st.root, h[i], rf[i]); 195 } 196 long long result = 0; 197 for(int i = 1; i <= n; i++) 198 smax(result, lf[i] + rf[i] - p[i]); 199 putInteger(result); 200 } 201 202 int main(){ 203 freopen("C.in", "r", stdin); 204 freopen("C.out", "w", stdout); 205 init(); 206 discretization(); 207 solve(); 208 return 0; 209 }