例题28 UVa10891 Game of Sum（DP）

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题意：

看白书

要点：

因为所以人都是用尽量拿最大值，所以用d[i][j]表示剩下序列为i~j时当前取的这个人得分的最大值，因为总共得分是固定的，一个人多另一个人就少，所以结果就是2*d(1,n)-sum(1,n)。可以写出状态转移方程为：d(i,j)=sum(i,j)-min{d(i+1,j)……d(j,j),d(i,j-1)……d(i,i),0}，这个式子的理解是：现在总共和为sum(i,j)，先手拿走的是总和减去剩下部分的另一个能达到的最大值中的最小值。

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int N = 100 + 10;int n;int a[N], s[N];int d[N][N],vis[N][N];int dp(int i, int j){if (vis[i][j])return d[i][j];vis[i][j] = 1;int ans = 0;for (int k = i + 1; k <= j; k++)ans = min(ans, dp(k,j));for (int k = i; k < j; k++)ans = min(ans, dp(i,k));d[i][j] = s[j] - s[i - 1] - ans;return d[i][j];}int main(){while (~scanf("%d", &n)&&n!=0){s[0] = 0;for (int i = 1; i <= n; i++){scanf("%d", &a[i]);s[i] = s[i - 1] + a[i];}memset(vis, 0, sizeof(vis));printf("%d\n", 2 * dp(1, n) - s[n]);}return 0;}

还可以通过递推将复杂都降到O(n^2)

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int N = 100 + 10;int n;int a[N], s[N];int f[N][N], g[N][N],d[N][N];int main(){while (~scanf("%d", &n)&&n!=0){s[0] = 0;for (int i = 1; i <= n; i++){scanf("%d", &a[i]);s[i] = s[i - 1] + a[i];}for (int i = 1; i <= n; i++)f[i][i] = g[i][i] = d[i][i] = a[i];for(int l=1;l<n;l++)for (int i = 1; i + l <= n; i++){int j = i + l;int m = 0;m = min(m, f[i + 1][j]);m = min(m, g[i][j - 1]);d[i][j] = s[j] - s[i - 1] - m;f[i][j] = min(d[i][j], f[i + 1][j]);g[i][j] = min(d[i][j], g[i][j - 1]);}printf("%d\n", 2 * d[1][n] - s[n]);}return 0;}

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