UVa 10891 Game of Sum (DP)

题意：给定一个长度为n的整数序列，两个人轮流从左端或者右端拿数，A先取，问最后A的得分-B的得分的结果。

析：dp[i][j] 表示序列 i~j 时先手得分的最大值，然后两种决策，要么从左端拿，要么从右端拿，肯定是拿的是最大的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#define debug() puts("++++");

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 100 + 5;

const int mod = 1e9 + 7;

const int dr[] = {-1, 1, 0, 0};

const int dc[] = {0, 0, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int dp[maxn][maxn], sum[maxn], a[maxn];

int f[maxn][maxn], g[maxn][maxn];

int main(){

    while(scanf("%d", &n) == 1 && n){

        for(int i = 1; i <= n; ++i)  scanf("%d", a+i);

        sum[0] = 0;

        for(int i = 1; i <= n; ++i){

            f[i][i] = g[i][i] = dp[i][i]  = a[i];

            sum[i] = sum[i-1] + a[i];

        }

        for(int l = 1; l < n; ++l)

            for(int i = 1; i+l <= n; ++i){

                int j = i + l;

                int mmin = 0;

                mmin = min(mmin, f[i+1][j]);

                mmin = min(mmin, g[i][j-1]);

                dp[i][j] = sum[j] - sum[i-1] - mmin;

                f[i][j] = min(dp[i][j], f[i+1][j]);

                g[i][j] = min(dp[i][j], g[i][j-1]);

            }

        printf("%d\n", 2*dp[1][n] - sum[n]);

    }

    return 0;

}