zoukankan html css js c++ java

（dp）HDU6199- gems gems gems

gems gems gems

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 476 Accepted Submission(s): 53

Problem Description

Now there are

Input

The first line contains an integer

Output

For each test case, print a single number in a line: the difference between the total value of gems Alice took and the total value of gems Bob took.

Sample Input

1 3 1 3 2

Sample Output

dp[i][j]表示以第i个数开始，当前先手选择连续j个的最大差值。无需考虑具体是哪个人操作，每个人都希望到自己时自己的值与对方的值差尽可能的大，故只需开二维即可。

转移方程为 dp[i][j]=min(-dp[i+j][j]+sum[i+j-1]-sum[i-1],-dp[i+j][j+1]+sum[i+j-1]-sum[i-1]) 其中有几个细节，一是如果i+j-1==n，则当前先手只有唯一选择dp[i][j]=sum[n]-sum[i-1] ，二是若i+j-1>n，则不存在（i，j）状态下先手的任何状态，三是递推时要保证（i+j,j）（i+j，j+1）如果算在转移中，一定要保证其为可行的先手状态。

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <vector>
  5 #include <set>
  6 #include <map>
  7 #include <string>
  8 #include <cstring>
  9 #include <stack>
 10 #include <queue>
 11 #include <cmath>
 12 #include <ctime>
 13 #include <bitset>
 14 #include <utility>
 15 #include <assert.h>
 16 using namespace std;
 17 #define rank rankk
 18 #define mp make_pair
 19 #define pb push_back
 20 #define xo(a,b) ((b)&1?(a):0)
 21 #define tm tmp
 22 //#define LL ll
 23 typedef unsigned long long ull;
 24 typedef pair<int,int> pii;
 25 typedef long long ll;
 26 typedef pair<ll,int> pli;
 27 typedef pair<ll,ll> pll;
 28 const int INF=0x3f3f3f3f;
 29 const ll INFF=0x3f3f3f3f3f3f3f3fll;
 30 const int MAX=2e6+5;
 31 const ll MAXN=2e8;
 32 const int MAX_N=MAX;
 33 const double da=2e9+5.0;
 34 const ll MOD=998244353;
 35 //const long double pi=acos(-1.0);
 36 //const double eps=0.00000001;
 37 int gcd(int a,int b){return b?gcd(b,a%b):a;}
 38 template<typename T>inline T abs(T a) {return a>0?a:-a;}
 39 template<class T> inline
 40 void read(T& num) {
 41     bool start=false,neg=false;
 42     char c;
 43     num=0;
 44     while((c=getchar())!=EOF) {
 45         if(c=='-') start=neg=true;
 46         else if(c>='0' && c<='9') {
 47             start=true;
 48             num=num*10+c-'0';
 49         } else if(start) break;
 50     }
 51     if(neg) num=-num;
 52 }
 53 inline ll powMM(ll a,ll b,ll M){
 54     ll ret=1;
 55     a%=M;
 56 //    b%=M;
 57     while (b){
 58         if (b&1) ret=ret*a%M;
 59         b>>=1;
 60         a=a*a%M;
 61     }
 62     return ret;
 63 }
 64 void open()
 65 {
 66 //    freopen("1009.in","r",stdin);
 67     freopen("out.txt","w",stdout);
 68 }
 69 ll dp[20005][150],sum[20005];
 70 int t,n,st;
 71 int main()
 72 {
 73     scanf("%d",&t);
 74     while(t--)
 75     {
 76         scanf("%d",&n);
 77         for(int i=1;i<=n;i++)scanf("%lld",&sum[i]),sum[i]+=sum[i-1];
 78         for(int i=1;i<=n;i++)
 79         {
 80             st=(int)ceil((sqrt(1.0+8.0*i)-1.0)/2.0)+1;
 81             for(int j=1;j<=st;j++)
 82                 dp[i][j]=INFF/2LL;
 83         }
 84         for(int i=n;i>=1;i--)
 85         {
 86             st=(int)ceil((sqrt(1.0+8.0*i)-1.0)/2.0)+1;
 87             for(int j=st;j>=1;j--)
 88             {
 89                 if(i+j-1==n)
 90                     dp[i][j]=sum[i+j-1]-sum[i-1];//选择唯一
 91                 else if(i+j-1>n)
 92                     dp[i][j]=INFF/2LL;//不存在
 93                 else
 94                 {
 95                     if(dp[i+j][j]==INFF/2LL&&dp[i+j][j]==INFF/2LL)dp[i][j]=sum[i+j-1]-sum[i-1];
 96                     else if(dp[i+j][j+1]==INFF/2LL)dp[i][j]=-dp[i+j][j]+sum[i+j-1]-sum[i-1];
 97                     else dp[i][j]=min(-dp[i+j][j]+sum[i+j-1]-sum[i-1],-dp[i+j][j+1]+sum[i+j-1]-sum[i-1]);//枚举后手的选择
 98                 }
 99             }
100         }
101         if(n==1)printf("%lld
",dp[1][1]);
102         else printf("%lld
",max(dp[1][1],dp[1][2]));
103     }
104 }

查看全文

相关阅读:
LuoguP1126 机器人搬重物（BFS）
POJ1950----DFS
C
B
A
C
B
A
A
B

原文地址：https://www.cnblogs.com/quintessence/p/7504089.html