USACO算法系列十八——game1

    题目：http://www.nocow.cn/index.php/Translate:USACO/game1

    很有趣的一道题目，两个人轮流从一串数字的两端拿数字，要求使自己拿到的数字总和最大。如果使用回溯的方法，限定不能保证最优的情况，即最优情况不知道怎么界定。

    可以用动态规划的思想来考虑，用MAP[i][j]表示，从i到j的数字串，自己能拿到的最大的数字和。那么有如下公式

    MAP[i][j] = sum[i][j] - min(map[i+1][j],map[i][j-1]); i 不等于j。（即，自己拿了一个数字以后，保证对方只能拿到比较少的。）

    MAP[i][j] = number[i] i == j时。

    sum[i][j]表示i 到 j，对number[]求和。

 #include <iostream>#include <fstream>#define SIZE 101using namespace std;int length;int number[SIZE] = {0};//SUM[I][J],I到J行的和int sum[SIZE][SIZE] = {0};//MAP[I][J],I到J行，最大选择值int map[SIZE][SIZE] = {0};//求left到right的和void sumarize(int left, int right){for (int i=left; i <= right; i ++){sum[left][right] += number[i];}}//求I到J行，最大选择值void game(int left, int right){if (left == right){map[left][right] = number[left];return;}else{if (map[left+1][right] == 0){game(left + 1, right);}if (map[left][right-1] == 0){game(left,right-1);}if (sum[left][right] == 0){sumarize(left,right);}map[left][right] = sum[left][right] - min(map[left+1][right],map[left][right-1]);}}int main(){ifstream fin("game1.in");ofstream fout("game1.out");//输入数据fin >> length;for (int i=1; i <= length; i ++){fin >> number[i];}game(1,length);fout << map[1][length] << " "<< sum[1][length] - map[1][length] << endl;return 0;}

    测试结果如下：

Executing... Test 1: TEST OK [0.000 secs, 3096 KB] Test 2: TEST OK [0.011 secs, 3096 KB] Test 3: TEST OK [0.000 secs, 3096 KB] Test 4: TEST OK [0.000 secs, 3096 KB] Test 5: TEST OK [0.011 secs, 3096 KB] Test 6: TEST OK [0.011 secs, 3096 KB] Test 7: TEST OK [0.011 secs, 3096 KB] Test 8: TEST OK [0.011 secs, 3096 KB] Test 9: TEST OK [0.011 secs, 3096 KB] Test 10: TEST OK [0.011 secs, 3096 KB] Test 11: TEST OK [0.011 secs, 3096 KB] Test 12: TEST OK [0.000 secs, 3096 KB] Test 13: TEST OK [0.000 secs, 3096 KB] Test 14: TEST OK [0.000 secs, 3096 KB] Test 15: TEST OK [0.011 secs, 3096 KB] Test 16: TEST OK [0.011 secs, 3096 KB]All tests OK.

     所以一定要注意思考题目的方式。要抽象，用数学的语言，最好是有规律的，明确的语言思考。