标签：exp private computed 最小 ast 位置 eric TE stream
problem1 link
最优选择一定是在$2n$个端点中选出两个。
problem2 link
分开考虑每个区间。设所有区间的左端点的最大值为$lc$,所有区间的右端点的最小值为$rc$.对于某个区间$L$，其实就是找最少的区间（包括$L$）能够完全覆盖区间$[lc,rc]$.
设$L$的左右端点为$L_{1},L_{2}$。考虑从$L_{1}$向左扩展，每次扩展一定是找一个区间$p$满足$p_{2}\geq L_{1}$且使得$p_{1}$最小。右端点向右扩展类似。
problem3 link
对于后手来说，它一定是每次只放置一个障碍格子在先手目前的位置。否则，如果放置了太多，那么先手就容易决策出向左还是向右比较好。
对于先手来说，假设目前先手在$(r,c)$，那么如果$(r,c)$和$(r+1,c)$之间没有障碍，且轮到先手，应该直接跳到$(r+1,c)$,否则，那么由后手的策略来看，此时已经有的障碍一定是一个包含$(r,c)$的区间，此时先手可以尝试向左或者向右运动到障碍区间外一个格子。
这样就可以用$(r,c,left,right,tag)$来表示一个状态来进行动态规划。$(r,c)$表示先手目前的位置，$(left, right)$表示目前障碍的区间，$tag$指示目前是先手还是后手。
code for problem1
code for problem2
#include <algorithm>
#include <string>
#include <vector>
constexpr int N = 2500;
constexpr int M = 10000;
int s[N], t[N];
int back[M], next[M];
int back_cost[M], next_cost[M];
int sort_idx[N];
class ShoutterDiv1 {
public:
int count(const std::vector<std::string>& s1000,
const std::vector<std::string>& s100,
const std::vector<std::string>& s10,
const std::vector<std::string>& s1,
const std::vector<std::string>& t1000,
const std::vector<std::string>& t100,
const std::vector<std::string>& t10,
const std::vector<std::string>& t1) {
int n = 0;
for (const auto& e : s1000) {
n += static_cast<int>(e.size());
}
for (int i = 0; i < n; ++i) {
s[i] = t[i] = 0;
sort_idx[i] = i;
}
for (int i = 0; i < M; ++i) {
back[i] = next[i] = i;
}
Parse(s1000, s, 1000);
Parse(s100, s, 100);
Parse(s10, s, 10);
Parse(s1, s, 1);
Parse(t1000, t, 1000);
Parse(t100, t, 100);
Parse(t10, t, 10);
Parse(t1, t, 1);
int c_left = t[0];
int c_right = s[0];
for (int i = 0; i < n; ++i) {
back[t[i]] = std::min(back[t[i]], s[i]);
next[s[i]] = std::max(next[s[i]], t[i]);
c_left = std::min(c_left, t[i]);
c_right = std::max(c_right, s[i]);
}
for (int i = M - 2; i >= 0; --i) {
back[i] = std::min(back[i], back[i + 1]);
}
for (int i = 1; i < M; ++i) {
next[i] = std::max(next[i], next[i - 1]);
}
for (int i = 0; i < M; ++i) {
if (i <= c_left) {
back_cost[i] = 0;
} else if (i == back[i]) {
back_cost[i] = -1;
} else {
if (back_cost[back[i]] == -1) {
back_cost[i] = -1;
} else {
back_cost[i] = 1 + back_cost[back[i]];
}
}
}
for (int i = M - 1; i >= 0; --i) {
if (i >= c_right) {
next_cost[i] = 0;
} else if (i == next[i]) {
next_cost[i] = -1;
} else {
if (next_cost[next[i]] == -1) {
next_cost[i] = -1;
} else {
next_cost[i] = 1 + next_cost[next[i]];
}
}
}
auto Cost = [&](int left, int right) {
if (back_cost[left] == -1 || next_cost[right] == -1) {
return -1;
}
return back_cost[left] + next_cost[right];
};
std::sort(sort_idx, sort_idx + n, [&](int x, int y) {
return s[x] < s[y] || (s[x] == s[y] && t[x] < t[y]);
});
int result = 0;
for (int i = 0; i < n; ++i) {
int tmp = Cost(s[sort_idx[i]], t[sort_idx[i]]);
for (int j = 0; j < N && s[sort_idx[j]] <= s[sort_idx[i]]; ++j) {
if (t[sort_idx[i]] <= t[sort_idx[j]]) {
int val = Cost(s[sort_idx[j]], t[sort_idx[j]]);
if (val != -1 && (tmp == -1 || tmp > 1 + val)) {
tmp = 1 + val;
}
}
}
if (tmp == -1) {
return -1;
}
result += tmp;
}
return result;
}
private:
void Parse(const std::vector<std::string>& d, int* data, int base) {
int idx = 0;
for (size_t i = 0; i < d.size(); ++i) {
for (size_t j = 0; j < d[i].size(); ++j) {
data[idx++] += (d[i][j] - ‘0‘) * base;
}
}
}
};
code for problem3
#include <cstring>
#include <limits>
#include <string>
#include <vector>
#include <iostream>
constexpr int N = 50;
constexpr int kEachUseBit = 16;
constexpr int kRabbitTagBit = 31;
constexpr int kEelTagBit = 15;
constexpr int kMask = (1 << (kEachUseBit - 1)) - 1;
using TType = unsigned int;
TType rabbit_eel[N][N][N][N + 1];
int cost[N][N];
int h, w;
class WallGameDiv1 {
public:
int play(const std::vector<std::string> &costs) {
h = static_cast<int>(costs.size());
w = static_cast<int>(costs[0].size());
for (int i = 0; i < h; ++i) {
cost[i][0] = costs[i][0] - ‘0‘;
for (int j = 1; j < w; ++j) {
cost[i][j] = cost[i][j - 1] + costs[i][j] - ‘0‘;
}
}
memset(rabbit_eel, 0, sizeof(rabbit_eel));
// Iterator each row to avoid more depths of recursion.
for (int i = h - 1; i >= 0; --i) {
for (int j = 0; j < w; ++j) {
Eel(i, j, j, j);
}
}
int result = std::numeric_limits<int>::max();
for (int i = 0; i < w; ++i) {
result = std::min(result, Cost(0, i, i) + Eel(0, i, i, i));
}
return result;
}
private:
bool Computed(const TType &val, bool is_rabbit) {
if (is_rabbit) {
return (val & (1u << kRabbitTagBit)) != 0;
} else {
return (val & (1u << kEelTagBit)) != 0;
}
}
int GetCost(const TType &val, bool is_rabbit) {
if (!Computed(val, is_rabbit)) {
return -1;
}
if (is_rabbit) {
return static_cast<int>((val >> kEachUseBit) & kMask);
} else {
return static_cast<int>(val & kMask);
}
}
int SetCost(TType &val, bool is_rabbit, int cost) {
if (is_rabbit) {
val |= (static_cast<TType>(cost)) << kEachUseBit;
val |= 1u << kRabbitTagBit;
} else {
val |= cost;
val |= 1u << kEelTagBit;
}
return cost;
}
int Eel(int row, int col, int left, int right) {
TType &val = rabbit_eel[row][col][left][right];
int result = GetCost(val, false);
if (result != -1) {
return result;
}
if (row == h - 1) {
return SetCost(val, false, 0);
}
result = Rabbit(row, col, left, right);
if (right - left < w - 1) {
result = std::max(result, Rabbit(row, col, std::min(col, left),
std::max(col + 1, right)));
}
return SetCost(val, false, result);
}
int Rabbit(int row, int col, int left, int right) {
TType &val = rabbit_eel[row][col][left][right];
int result = GetCost(val, true);
if (result != -1) {
return result;
}
if ((left <= col) && (col < right)) {
result = std::numeric_limits<int>::max();
if (left > 0) {
result = Cost(row, left - 1, col - 1) + Eel(row, left - 1, left, right);
}
if (right < w) {
result = std::min(
result, Cost(row, col + 1, right) + Eel(row, right, left, right));
}
} else {
result = Cost(row + 1, col, col) + Eel(row + 1, col, col, col);
}
return SetCost(val, true, result);
}
int Cost(int row, int left, int right) {
if (left <= right) {
if (left == 0) {
return cost[row][right];
}
return cost[row][right] - cost[row][left - 1];
}
return 0;
}
};
标签：exp private computed 最小 ast 位置 eric TE stream
原文地址：https://www.cnblogs.com/jianglangcaijin/p/9250256.html
