Financiers Game CodeForces - 737D (博弈论,区间dp)

大意: 给定$n$元素序列, 两个人从两端轮流拿数, 每一步假设对手上次取k, 那么只能取k或k+1, 先手第一步取1或2, 直到不能拿时停止. 先手要最大化两人数字和的差, 后手要最小化, 求最后差是多少.

显然状态数是$O(n^2)$的, 直接暴力DP

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head

const int N = 4010;
int n, a[N];
int *dp[2][N][N];
int o;
int dfs(int tp, int l, int r, int k) {
	if (!dp[tp][l][r]) {
		int *p = new int[100];
		memset(p,0x3f,sizeof(int)*100);
		dp[tp][l][r] = p;
	}
	int &ans = dp[tp][l][r][k];
	if (ans!=INF) return ans;
	if (r-l+1<k) return ans=0;
	if (r-l+1==k) return ans=tp?a[r]-a[l-1]:a[l-1]-a[r];
	if (tp==1) return ans=max(dfs(0,l+k,r,k)+a[l+k-1]-a[l-1],dfs(0,l+k+1,r,k+1)+a[l+k]-a[l-1]);
	return ans=min(dfs(1,l,r-k,k)-a[r]+a[r-k],dfs(1,l,r-k-1,k+1)-a[r]+a[r-k-1]);
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i),a[i]+=a[i-1];
	printf("%d\n", dfs(1,1,n,1));
}

原文地址：https://www.cnblogs.com/uid001/p/10620350.html