P1880 [NOI1995]石子合并 区间dp

P1880 [NOI1995]石子合并

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 const int inf = 0x3f3f3f3f;
 int cost1[maxn][maxn], cost2[maxn][maxn];  //当前合并的代价
 int dp1[maxn][maxn], dp2[maxn][maxn];
 int main() {
     int n; cin >> n;
     for (int i = ; i <= n; i++) {
         cin >> cost1[i][i];
         cost2[i][i] = cost1[i][i];
     }
     for (int i = ; i <= n; i++) {
         cost1[n+i][n+i] = cost1[i][i];
         cost2[n+i][n+i] = cost2[i][i];
     }
     n <<= ;
     for (int i = ; i <= n/; i++) {
         for (int j = ; j <= n-i; j++) {
             int x = i+j, y = j+;
             for (int k = ; k < i; k++) {
                 cost1[x][y] = cost1[x-k][y] + cost1[x][y+i-k];
                 cost2[x][y] = cost2[x-k][y] + cost2[x][y+i-k];
                 if (dp1[x][y]) dp1[x][y] = min(dp1[x][y],cost1[x][y]+dp1[x-k][y]+dp1[x][y+i-k]);
                 else dp1[x][y] = cost1[x][y]+dp1[x-k][y]+dp1[x][y+i-k];
                 if (dp2[x][y]) dp2[x][y] = max(dp2[x][y],cost2[x][y]+dp2[x-k][y]+dp2[x][y+i-k]);
                 else dp2[x][y] = cost2[x][y]+dp2[x-k][y]+dp2[x][y+i-k];
             }
         }
     }
     int mi = inf, mx = ;
     n >>= ;
     for(int j = ; j <= n; j++) {
         mi = min(mi,dp1[n+j][j+]);
     }
     for(int j = ; j <= n; j++) {
         mx = max(mx,dp2[n+j][j+]);
     }
     cout << mi << endl << mx << endl;
 }