sicily 1176 Two Ends dp(记忆化搜索)

今天在网络中心上班，看到他们在做这题，吃饭后我也来做，刚开始看，就直接想暴力dfs，看了一下status，130个TLE，断定不行了，哈哈，我算法分析不行，只能这样看。

然后想了一下，其实也是最优问题，dp可以解决，不过还是WA了一次，因为没留意When employing the greedy strategy, always take the larger end. If there is a tie, remove the left end.这个条件。加等号就A了

#include <iostream> #include <cstdio> #include <cstring> using namespace std; int n; int num[1001]; int sum; int f[1001][1001]; int dp(int beg, int end) { if(f[beg][end] == 0) { if(beg == end - 1) //剩下两个数的时候真的要贪心了 return f[beg][end] = max(num[beg], num[end]); //分情况讨论 int tmp1 = num[beg]; if(num[beg + 1] >= num[end]) //注意等号，如果两个边相等，则取左边 tmp1 += dp(beg + 2, end); else tmp1 += dp(beg + 1, end - 1); int tmp2 = num[end]; if(num[beg] >= num[end - 1]) tmp2 += dp(beg + 1, end - 1); else tmp2 += dp(beg, end - 2); f[beg][end] = max(tmp1, tmp2); return f[beg][end]; } return f[beg][end]; } int main() { //freopen("1.txt", "r", stdin); int count = 1; while(scanf("%d", &n) != EOF && n != 0) { sum = 0; memset(f, 0, sizeof(f)); for(int i = 0; i < n; i++) { scanf("%d", &num[i]); sum += num[i]; } printf("In game %d, the greedy strategy might lose by as many as %d points./n", count++, 2 * dp(0, n - 1) - sum); //得分相减 } return 0; }