51nod1489 (DFS)

Title Link: https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1489

Test instructions: Chinese question eh ~

Idea: DFS

First we will kill the first and last person by attacking the 1th and the 2nd person.

Next we consider how to kill the middle man most efficiently:

Assuming that we are attacking the POS person at the moment, then we have to make sure that the pos-1 person is dead to move down a person (this is well understood, because we cannot kill it if we have not killed the first pos-1 person before attacking the POS person). But we need to be more specific about these situations:

1. If we die at the point where the POS person dies, the pos person is dead and we have not killed the pos-1 person, then we need to continue attacking the POS individual until pos-1 death (because we are using DFS, so there is no need to consider returning to direct attack on the pos-1 person);

2. If we attack the POS person, POS is not dead, pos-1 is dead, assuming we attack CNT kill pos-1, Attack cc times can kill Pos, then we need to traverse the attack cnt~cc attacks into the next level of recursion all the circumstances;

The best solution we've got is the answer.

Code:

1#include <bits/stdc++.h>2 #defineMAXN 503 #defineINF 0x3f3f3f3f4 using namespacestd;5 6 intGG[MAXN], ans=INF, n, a, B;7 8 voidDfsintPosintnum) {9     intCnt=0, cc=0;//* * * Note here the CNT must be assigned an initial value, because it is possible to meet the (cc>cnt&&gg[pos]>=0) condition when the condition is not satisfied (gg[pos-1]>=0)Ten     if(pos==n) {//* * * to reach nth person is to return and update the number of attacks Oneans=min (ans, num); A         return; -     } -     if(gg[pos-1]<0){//* * * Only the previous person is dead to move backwards theDFS (pos+1, num); -     } -     if(gg[pos-1]>=0){//* * * Handling only cases greater than or equal to 0 -cnt=gg[pos-1]/b+1; +gg[pos-1]-=b*CNT; -gg[pos]-=a*CNT; +gg[pos+1]-=b*CNT; ADFS (pos+1, num+CNT);  atgg[pos-1]+=b*cnt;//* * * pay attention to backtracking to restore the amount of the previous change -gg[pos]+=a*CNT; -gg[pos+1]+=b*CNT; -     } -cc=gg[pos]/a+1; -     if(cc>cnt&&gg[pos]>=0){//* * * If pos-1 is dead at this point and POS is still alive, we can choose to kill it directly and move it to the next one or move it directly down a in          for(inti=cnt+1; i<=cc; i++){ -gg[pos-1]-=b*i; togg[pos]-=a*i; +gg[pos+1]-=b*i; -DFS (pos+1, num+i); thegg[pos-1]+=b*i; *gg[pos]+=a*i; $gg[pos+1]+=b*i;Panax Notoginseng         } -     } the     return; + } A  the intMainvoid){ +     intnum=0; -CIN >> N >> a >>b; $      for(intI=1; i<=n; i++){ $CIN >>Gg[i]; -     } -     intcnt=gg[1]/b+1;//* * * First go head thenum+=CNT; -gg[1]-=b*CNT;Wuyigg[2]-=a*CNT; thegg[3]-=b*CNT; -     if(gg[n]>=0){//* * * to the end, note that if only three people can be gg[n] this time is less than 0 Wucnt=gg[n]/b+1; -num+=CNT; Aboutgg[n-2]-=b*CNT; $gg[n-1]-=a*CNT; -gg[n]-=b*CNT; -     } -Dfs2,0); A     if(Ans==inf) {//* * * Note there is a possibility that DFS directly satisfies the condition exit, at which time the ANS value is INF +ans=0; the     } -cout << Ans+num <<Endl; $     return 0; the}

51nod1489 (DFS)