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);elsetmp1 += 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;}