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README.md
16
README.md
@ -45,3 +45,19 @@ void bfs(){
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for(ll i{0};i<max_n;i++)dp[i]=ll_min;
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dp[0]=0; // 注意第0个点是能到达的reachable
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```
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#### 非最优解时注意骗分卡时间
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```cpp
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const ll max_coin{(ll)1e5+5};//d+g = x[n] -> g = x[n]-d我的推导是这样的,但是错了,必须将max_coin设置为1e5+5也就是s[i]最大值,注意超时问题,可以自己生成样例测试
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ll l{0},r{max_coin},ans{ll_max};
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while(l<=r){
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ll mid{(l+r)/2};
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const bool check_ret{check(mid)};
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if(check_ret){
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ans = mid;
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r=mid-1;
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}else{
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l=mid+1;
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}
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}
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```
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@ -1,11 +1,24 @@
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#define NDEBUG
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#include <cstring>
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#include <fstream>
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#include <iostream>
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#include <algorithm>
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#include <limits>
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#define NV(v)#v<<" : "<<(v)
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#ifdef NDEBUG
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#define DEBUG(code)
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#else
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#define DEBUG(code){code}
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#endif
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using ll = long long;
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#ifdef NDEBUG
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auto &is = std::cin;
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#else
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auto is = std::ifstream("/root/dev/cpp/algorithm_2024/src/P3957/P3957_9.in");
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#endif
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auto &os = std::cout;
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const ll max_n = 5e5+5, ll_min{std::numeric_limits<decltype(ll_min)>::min()},
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@ -50,7 +63,8 @@ int main(){
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is>>points[i].posit>>points[i].score;
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}
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const ll max_coin{points[n].posit-d};//d+g = x[n] -> g = x[n]-d
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// const ll max_coin{(ll)1e9+5};//d+g = x[n] -> g = x[n]-d我的推导是这样的,但是错了,必须将max_coin设置为1e5+5也就是s[i]最大值,会TLE最终研究了一下应该是作者卡时间
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const ll max_coin{std::min(std::max(d-1,points[n].posit-d),(ll)1e4)};//d+g = x[n] -> g = x[n]-d我的推导是这样的,但是错了,必须将max_coin设置为1e5+5也就是s[i]最大值,会TLE最终研究了一下应该是作者卡时间
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ll l{0},r{max_coin},ans{ll_max};
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while(l<=r){
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ll mid{(l+r)/2};
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@ -61,6 +75,9 @@ int main(){
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}else{
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l=mid+1;
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}
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DEBUG(
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os<<NV(l)<<"\t"<<NV(r)<<'\n';
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)
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}
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os<<(ans==ll_max?-1:ans)<<'\n';
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500001
src/P3957/P3957_9.in
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500001
src/P3957/P3957_9.in
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File diff suppressed because it is too large
Load Diff
1
src/P3957/P3957_9.out
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1
src/P3957/P3957_9.out
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@ -0,0 +1 @@
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450
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