UVa 10891 Game of Sum - 动态规划

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　　因为数的总和一定，所以用一个人得分越高，那么另一个人的得分越低。　　

　　用$dp[i][j]$表示从$[i, j]$开始游戏，先手能够取得的最高分。

　　转移通过枚举取的数的个数$k$来转移。因为你希望先手得分尽量高，所以另一个人的最高得分应尽量少。

　　$dp[i][j] = sum[i][j] - \min \{dp[i + k][j],dp[i][j - k]\}$

　　但是发现计算$dp[i + k][j],dp[i][j - k]$的最小值的地方很重复，所以用一个$f[i][j]$储存前者的最优值，$g[i][j]$储存后者的最优值。

　　这样就将代码的时间复杂度优化到O(n²)

Code

 /**
  * uva
  * Problem#10891
  * Accepted
  * Time:0ms
  */
 #include<iostream>
 #include<cstdio>
 #include<cctype>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<sstream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 #include<stack>
 using namespace std;
 typedef bool boolean;
 #define INF 0xfffffff
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 template<typename T>
 inline void readInteger(T& u){
     char x;
     long long aFlag = ;
     while(!isdigit((x = getchar())) && x != '-');
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
 }

 int n;
 int *list;
 int f[][];
 int g[][];
 int dp[][];

 inline boolean init(){
     readInteger(n);
     if(n == )    return false;
     list = new int[(const int)(n + )];
     for(int i = ; i <= n; i++){
         readInteger(list[i]);
     }
     return true;
 }

 int *sum;
 inline void getSum(){
     sum = new int[(const int)(n + )];
     sum[] = ;
     for(int i = ; i <= n; i++)
         sum[i] = sum[i - ] + list[i];
 }

 inline void solve(){
     memset(f, 0x7f, sizeof(f));
     memset(g, 0x7f, sizeof(g));
     for(int i = ; i <= n; i++)    f[i][i] = g[i][i] = dp[i][i] = list[i];
     for(int k = ; k < n; k++){
         for(int i = ; i + k <= n; i++){
             int j = i + k;
             int m = ;
             smin(m, f[i + ][j]);
             smin(m, g[i][j - ]);
             dp[i][j] = sum[j] - sum[i - ] - m;
             f[i][j] = min(f[i + ][j], dp[i][j]);
             g[i][j] = min(g[i][j - ], dp[i][j]);
         }
     }
     printf("%d\n", dp[][n] *  - sum[n]);
     delete[] list;
     delete[] sum;
 }

 int main(){
     while(init()){
         getSum();
         solve();
     }
     return ;
 }