UVA - 10891 - Game of Sum （DP）

题目传送：UVA - 10891

思路：定义dp(i,j)表示原序列中的第i~j个元素组成的子序列，在双方都采取最优策略的情况下，先手得分的最大值

通过枚举给对方剩下怎样的子序列，有

状态转移方程为：dp(i, j) = sum(i, j) - min{d(i+1, j), d(j, j) ,d(i,j-1),...,d(i,i),0};

AC代码①(On^3)：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 105;
int d[maxn][maxn];
int vis[maxn][maxn], a[maxn]; 
int sum[maxn];
int n;

int dp(int i, int j) {
	if(vis[i][j]) return d[i][j];
	vis[i][j] = 1;
	int m = 0;
	for(int k = i; k < j; k ++) m = min(m, dp(i, k));
	for(int k = i + 1; k <= j; k ++) m = min(m, dp(k, j));
	d[i][j] = sum[j] - sum[i - 1] - m;
	return d[i][j];
}

int main() {
	while(scanf("%d", &n) != EOF) {
		if(n == 0) break;
		sum[0] = 0;
		for(int i = 1; i <= n; i ++) {
			scanf("%d", &a[i]);
			sum[i] = sum[i-1] + a[i];
		}
		memset(vis, 0, sizeof(vis));
		printf("%d\n", 2 * dp(1, n) - sum[n]);
	}
	return 0;
}

AC代码②(On^2)：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 105;
int d[maxn][maxn];//记录先手在区间i到j中取得的最大值 
int f[maxn][maxn];//记录min{d(i,j),d(i+1,j),...,d(j,j)},后缀 
int g[maxn][maxn];//记录min{d(i,j),d(i,j-1),...,d(i,i)},前缀 
int a[maxn];
int sum[maxn];
int n;

int main() {
	while(scanf("%d", &n) != EOF) {
		if(n == 0) break;
		sum[0] = 0;
		for(int i = 1; i <= n; i ++) {
			scanf("%d", &a[i]);
			sum[i] = sum[i-1] + a[i];
		}
		for(int i = 1; i <= n; i ++) {
			f[i][i] = g[i][i] = d[i][i] = a[i];
		}
		for(int len = 1; len <= n; len ++) {
			for(int i = 1; i + len <= n; i ++) {
				int j = i + len;
				int m = 0; 
				m = min(m, f[i + 1][j]);
				m = min(m, g[i][j - 1]);
				d[i][j] = sum[j] - sum[i - 1] - m;
				f[i][j] = min(d[i][j], f[i+1][j]);
				g[i][j] = min(d[i][j], g[i][j-1]);
			}
		}
		cout << 2 * d[1][n] - sum[n] << endl;
	}
	return 0;
}