LightOJ 1031 Easy Game(区间DP)

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题意：A，B轮流从左或从右以最优策略选出连续的几个数，直至选完，求两人所选数和差的最小值。

思路：dp[i][j]表示先手者A选区间[i, j]的最优策略和。A要让B得到的和最小，所以先求出让B最优策略得到的最小数字和，即 tmp = min(dp[i][k], dp[k+1][j])，其中 i <= k < j。这样得到tmp为可以让B最优策略最小的值，则有 dp[i][j] = sum(i, j) - tmp。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int a[N], sum[N];int dp[N][N];int min(int a, int b, int c) { return min(a, min(b, c)); }int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    si(n);    mem(dp, 0), mem(sum, 0);    for(int i = 1; i <= n; i ++) si(a[i]), sum[i] = sum[i-1] + a[i];    for(int j = 1; j <= n; j ++) {    for(int i = j; i > 0; i --) {    int tmp = 0;    for(int k = i; k < j; k ++)    tmp = min(tmp, dp[i][k], dp[k+1][j]);    dp[i][j] = sum[j] - sum[i-1] - tmp;      }    }    printf("Case %d: %d\n", ++ cas, 2 * dp[1][n] - sum[n]);    }        return 0;}

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