矩阵拿宝物--Codeforces 1201D - Treasure Hunting Codeforces Round #577 (Div. 2)
网上题解比较少，自己比较弱研究了半天（已经过了），希望对找题解的人有帮助
题目链接：https://codeforc.es/contest/1201/problem/D
题意: 给你一个矩形，起始点在(1,1),在给定坐标有宝物，你要将整个图中的宝物全部拿到，而且你不能向下走（左右随意），而且只有在所给出的列上你才能向上走，问最少要走多少格
#define IOS ios_base::sync_with_stdio(0); cin.tie(0); #include <cstdio>//sprintf islower isupper #include <cstdlib>//malloc exit strcat itoa system("cls") #include <iostream>//pair #include <fstream> #include <bitset> //#include <map> //#include<unordered_map> https://codeforc.es/contest/1201/problem/D #include <vector> #include <stack> #include <set> #include <string.h>//strstr substr #include <string> #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; #include <cmath> #include <deque> #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less #include <vector>//emplace_back //#include <math.h> //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation #define fo(a,b,c) for(register int a=b;a<=c;++a) #define fr(a,b,c) for(register int a=b;a>=c;--a) #define mem(a,b) memset(a,b,sizeof(a)) #define pr printf #define sc scanf void swapp(int &a,int &b); double fabss(double a); int maxx(int a,int b); int minn(int a,int b); int Del_bit_1(int n); int lowbit(int n); int abss(int a); const long long INF=(1LL<<60); const double E=2.718281828; const double PI=acos(-1.0); const int inf=(1<<29); const double ESP=1e-9; const int mod=(int)1e9+7; const int N=(int)2e5+10; int dis(int a,int b) { return abss(a-b); } int L[N],R[N],LV[N],RV[N]; struct node { int x,y; friend bool operator<(node a,node b) { if(a.y==b.y) return a.x<b.x; return a.y<b.y; } }_[N]; int is[N]; int main() { int n,m,k,q,n_=0; sc("%d%d%d%d",&n,&m,&k,&q); fo(i,1,k)sc("%d%d",&_[i].y,&_[i].x);//题目中取完一行只可能停留在头和尾 sort(_+1,_+1+k); //下一行进行转移就行了 fo(i,1,k) { int tx=_[i].x,ty=_[i].y; n_=maxx(n_,ty); if(!LV[ty])LV[ty]=tx; //预处理 else LV[ty]=minn(tx,LV[ty]);//宝藏的最左 if(!RV[ty])RV[ty]=tx; else RV[ty]=maxx(tx,RV[ty]);//宝藏的最右 } fo(i,1,q)sc("%d",&is[i]),L[is[i]]=R[is[i]]=is[i]; sort(is+1,is+1+q); //预处理 fo(i,1,m)if(!L[i])L[i]=L[i-1];//左邻近的安全列 fr(i,m,1)if(!R[i])R[i]=R[i+1];//右邻近的安全列 int posl=1,posr=1; long long ans=0,preresl=0,preresr=0; fo(i,1,n_) { if(i==1) { if(RV[i]) ans=dis(posl,RV[i]),posl=posr=RV[i]; preresl=preresr=ans; } else { preresl++;preresr++;//除了第一行其他只要上升了,用于转移的左右停留状态都要+1; if(LV[i]==0&&RV[i]==0) continue; long long resl=INF,resr=INF; if(L[posl])resl=min(resl,preresl+dis(posl,L[posl])+dis(L[posl],RV[i])+dis(RV[i],LV[i])); if(R[posl])resl=min(resl,preresl+dis(posl,R[posl])+dis(R[posl],RV[i])+dis(RV[i],LV[i])); if(L[posr])resl=min(resl,preresr+dis(posr,L[posr])+dis(L[posr],RV[i])+dis(RV[i],LV[i])); if(R[posr])resl=min(resl,preresr+dis(posr,R[posr])+dis(R[posr],RV[i])+dis(RV[i],LV[i])); //进行转移; if(L[posl])resr=min(resr,preresl+dis(posl,L[posl])+dis(L[posl],LV[i])+dis(RV[i],LV[i])); if(R[posl])resr=min(resr,preresl+dis(posl,R[posl])+dis(R[posl],LV[i])+dis(RV[i],LV[i])); if(L[posr])resr=min(resr,preresr+dis(posr,L[posr])+dis(L[posr],LV[i])+dis(RV[i],LV[i])); if(R[posr])resr=min(resr,preresr+dis(posr,R[posr])+dis(R[posr],LV[i])+dis(RV[i],LV[i])); ans=min(resl,resr); preresl=resl; preresr=resr; posl=LV[i]; posr=RV[i]; } } pr("%lld\n",ans); return 0; } /**************************************************************************************/ int maxx(int a,int b) { return a>b?a:b; } void swapp(int &a,int &b) { a^=b^=a^=b; } int lowbit(int n) { return n&(-n); } int Del_bit_1(int n) { return n&(n-1); } int abss(int a) { return a>0?a:-a; } double fabss(double a) { return a>0?a:-a; } int minn(int a,int b) { return a<b?a:b; }
