uva 10891 Game of Sum (DP水题)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1832

　　博弈类的DP，通过枚举较短一段的最大得分推出较长一段的最大得分。转移方程是 dp[i][j] = sum[i][j] - min(dp[i][i], dp[i][i + 1], ... , dp[i][j - 1], dp[i + 1][j], dp[i + 2][j], ... ,dp[j][j])。

O(n^3)算法：

View Code

 1 #define REP(i, n) for (int i = 0; i < (n); i++)
 2 #define REP_1(i, n) for (int i = 1; i <= (n); i++)
 3 
 4 int dp[N][N], sum[N];
 5 
 6 int main() {
 7     int n;
 8     while (~scanf("%d", &n) && n) {
 9         sum[0] = 0;
10         REP_1(i, n) {
11             scanf("%d", &sum[i]);
12             sum[i] += sum[i - 1];
13         }
14         _clr(dp);
15         REP_1(d, n) {
16             REP_1(i, n - d + 1) {
17                 int j = i + d - 1, tmp = sum[j] - sum[i - 1];
18                 dp[i][j] = tmp;
19                 REP(l, d - 1) {
20                     dp[i][j] = max(dp[i][j], tmp - dp[i][i + l]);
21                     dp[i][j] = max(dp[i][j], tmp - dp[j - l][j]);
22                 }
23             }
24         }
25 //        printf("%d %d\n", dp[1][n], sum[n] - dp[1][n]);
26         printf("%d\n", (dp[1][n] << 1) - sum[n]);
27     }
28     return 0;
29 }

在计算较短段的时候记录头尾的最小值，可以得到O(n^2)算法：

View Code

 1 #define REP(i, n) for (int i = 0; i < (n); i++)
 2 #define REP_1(i, n) for (int i = 1; i <= (n); i++)
 3 
 4 int dp[N][N], sum[N], rec[2][N];
 5 
 6 int main() {
 7     int n;
 8     while (~scanf("%d", &n) && n) {
 9         sum[0] = 0;
10         REP_1(i, n) {
11             scanf("%d", &sum[i]);
12             sum[i] += sum[i - 1];
13         }
14         _clr(dp);
15         REP(i, 2) REP_1(j, n) rec[i][j] = inf;
16         REP_1(d, n) {
17             REP_1(i, n - d + 1) {
18                 int j = i + d - 1, tmp = sum[j] - sum[i - 1];
19                 dp[i][j] = max(tmp, max(tmp - rec[0][i], tmp - rec[1][j]));
20                 rec[0][i] = min(dp[i][j], rec[0][i]);
21                 rec[1][j] = min(dp[i][j], rec[1][j]);
22             }
23         }
24 //        printf("%d %d\n", dp[1][n], sum[n] - dp[1][n]);
25         printf("%d\n", (dp[1][n] << 1) - sum[n]);
26     }
27     return 0;
28 }

——written by Lyon

posted @ 2013-02-18 01:05 LyonLys 阅读(217) 评论(0) 收藏举报

刷新页面返回顶部