luogu-p1095 Watchman's escape--separate DP

P1095 Watchman's escape: https://www.luogu.org/problemnew/show/P1095

Test instructions

There is a person to run in a straight line of s length, the initial magic value of M, with 10 mana can run 60 meters in a second, and ordinary run a second 17 meters. You can restore the Mana value of 4 points by keeping still. Ask if you can run out of T-seconds ago.

Ideas:

Separated two times DP, first run out can use acceleration to use the speed of the journey. Compare the values of DP "I" and DP "I-1" +17 for the second time.

#include <algorithm>#include<iterator>#include<iostream>#include<cstring>#include<iomanip>#include<cstdlib>#include<cstdio>#include<string>#include<vector>#include<bitset>#include<cctype>#include<queue>#include<cmath>#include<list>#include<map>#include<Set>//#include <unordered_map>//#include <unordered_set>//#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/hash_policy.hpp>using namespacestd;//#pragma GCC optimize (3)//#pragma COMMENT (linker, "/stack:102400000,102400000")//C + +#defineLson (L, Mid, RT << 1)#defineRson (mid + 1, R, RT << 1 | 1)#defineDebug (x) cerr << #x << "=" << x << "\ n";#definePB Push_back#definePQ Priority_queuetypedefLong Longll;typedef unsignedLong LongUll;typedef pair<ll, LL >Pll;typedef pair<int,int>Pii;typedef pair<int,pii>P3;//priority_queue<int> Q;//This is a big root heap Q//priority_queue<int,vector<int>,greater<int> >q;//This is a little Gan q//__gnu_pbds::cc_hash_table<int,int>ret[11]; //It's a quick hash_map.#defineFi first#defineSe Second//#define Endl ' \ n '#defineOKC Ios::sync_with_stdio (false); Cin.tie (0)#defineFT (A,B,C) for (int a=b; A <= C;++a)//used to press the line#defineREP (I, J, K) for (int i = j; i < K; ++i)//Priority_queue<int, Vector<int>, greater<int> >que;Constll mos = 0X7FFFFFFFLL;//2147483647Constll nmos = 0X80000000LL;//-2147483648Const intINF =0x3f3f3f3f;Constll inff = 0X3F3F3F3F3F3F3F3FLL;// -Const DoublePi=acos (-1.0); template<typename t>inline T read (t&x) {x=0;intf=0;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') f|= (ch=='-'), ch=GetChar ();  while(ch>='0'&&ch<='9') x=x*Ten+ch-'0', ch=GetChar (); returnx=f?-x:x;}/*-----------------------Showtime----------------------*/            Const intMAXN =300009; intdp[maxn],m,s,t;intMain () {CIN>>m>>s>>T;  for(intI=1; i<=t; i++){                if(M >=Ten) {Dp[i]= dp[i-1]+ -; M-=Ten; }                Else{Dp[i]= dp[i-1]; M+=4; }            }             for(intI=1; i<=t; i++) {Dp[i]= Max (Dp[i], dp[i-1]+ -); if(dp[i]>=s) {Puts ("Yes"); printf ("%d\n", i); return 0; }} puts ("No"); printf ("%d\n", dp[t]); return 0;}

P1095

luogu-p1095 Watchman's escape--separate DP