UVA 10891 Game of Sum 总和一定的博弈，区间dp

参考:刘汝佳大白书p67-p69

整数的总和是一定的，所以一个人得分越高，另一个人的得分越低。

不管任意时刻游戏的状态都是原始序列的一段连续子序列，（因而可用区间dp）

//#pragma warning (disable: 4786)//#pragma comment (linker, "/STACK:16777216")//HEAD#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)typedef long long LL;const int INF = 1000000007;const double eps = 1e-10;const int MAXN = 110;int n, dp[MAXN][MAXN];int a[MAXN];int sum[MAXN];bool vis[MAXN][MAXN];int Left[MAXN][MAXN], Right[MAXN][MAXN];//(1)n^3//int dpf(int l, int r)//{//    if (l == r) return a[l];//    if (vis[l][r]) return dp[l][r];//    vis[l][r] = true;//    int &ans  = dp[l][r];//    ans = sum[r] - sum[l - 1];//    FE(i, l, r - 1)//    {////        ans = max(ans, sum[i] - sum[l - 1] + sum[r] - sum[i] - dpf(i + 1, r));//        ans = max(ans, sum[r] - sum[l - 1] - dpf(i + 1, r));//    }//    FED(i, r, l + 1)//    {////        ans = max(ans, sum[r] - sum[i - 1] + sum[i - 1] - sum[l - 1] - dpf(l, i - 1));//        ans = max(ans, sum[r] - sum[l - 1] - dpf(l, i - 1));//    }////    return ans;//}//(2)n^3//int dpf(int l, int r)//{//    if (l == r) return a[l];//    if (vis[l][r]) return dp[l][r];//    vis[l][r] = true;//    int tmp = 0;////    FE(i, l, r - 1)//    {//        tmp = min(tmp, dpf(i + 1, r));//    }//    FED(i, r, l + 1)//    {//        tmp = min(tmp, dpf(l, i - 1));//    }//    dp[l][r] = sum[r] - sum[l - 1] - tmp;//    return dp[l][r];//}//(3)n^2int solve(){    FE(i, 1, n)    {        Left[i][i] = Right[i][i] = dp[i][i] = a[i];    }    for (int L = 1; L < n; L++)///长度    {        for (int i = 1; i + L <= n; i++)        {            int j = i + L;///区间为[i, j]            int tmp = 0;            tmp = min(tmp, Left[i + 1][j]);            tmp = min(tmp, Right[i][j - 1]);            dp[i][j] = sum[j] - sum[i - 1] - tmp;            Left[i][j] = min(dp[i][j], Left[i + 1][j]);            Right[i][j] = min(dp[i][j], Right[i][j - 1]);        }    }    return dp[1][n];}int main (){    while (~RI(n) && n)    {        sum[0] = 0;        FE(i, 1, n)        {            RI(a[i]);            sum[i] = sum[i - 1] + a[i];        }        printf("%d\n", solve() * 2 - sum[n]);//        CLR(vis, 0);//        printf("%d\n", dpf(1, n) * 2 - sum[n]);    }    return 0;}