这样的状态变换还是第一次做,题意说:给你一个n长的递增的序列,要求分成k个序列,保证分成的序列最大的那部分和要最小。
状态方程:
dp[i][j] 表示前j个元素分成i组对应序列最大部分的和最小。 dp[i][j] = min(dp[i][j], max(dp[i - 1][k], sum[j] - sum[k]));
然后从前往后正常输出就好注意最后一个位置不用输出'/'
#include<iostream> #include<math.h> #include<stdio.h> #include<algorithm> #include<string.h> #include<string> #include<vector> #include<queue> #include<map> #include<set> #include<stack> #define B(x) (1<<(x)) using namespace std; typedef long long ll; typedef unsigned long long ull; typedef unsigned ui; const int oo = 0x3f3f3f3f; //const ll OO = 0x3f3f3f3f3f3f3f3f; const double eps = 1e-9; #define lson rt<<1 #define rson rt<<1|1 void cmax(int& a, int b){ if (b > a)a = b; } void cmin(int& a, int b){ if (b < a)a = b; } void cmax(ll& a, ll b){ if (b > a)a = b; } void cmin(ll& a, ll b){ if (b < a)a = b; } void cmax(double& a, double b){ if (a - b < eps) a = b; } void cmin(double& a, double b){ if (b - a < eps) a = b; } void add(int& a, int b, int mod){ a = (a + b) % mod; } void add(ll& a, ll b, ll mod){ a = (a + b) % mod; } const ll MOD = 1000000007; const int maxn = 510; int dp[maxn][maxn]; int a[maxn], sum[maxn], b[maxn]; int main(){ //freopen("E:\\read.txt", "r", stdin); int T, n, k; scanf("%d", &T); while (T--){ scanf("%d %d", &n, &k); for (int i = 1; i <= n; i++){ scanf("%d", &a[i]); sum[i] = sum[i - 1] + a[i]; } memset(dp, 0x3f, sizeof dp); memset(b, 0, sizeof b); for (int i = 1; i <= n; i++) dp[1][i] = sum[i]; for (int i = 2; i <= k; i++){ for (int j = i; j <= n; j++){ for (int k = i - 1; k < j; k++){ cmin(dp[i][j], max(dp[i - 1][k], sum[j] - sum[k])); } } } int x = 0, y = k - 1; for (int i = n; i >= 1; i--){ x += a[i]; if (x > dp[k][n]){ x = 0; y--; b[i] = 1; i++; } if (i <= y){ for (; i >= 1; i--) b[i] = 1; break; } } //printf("%d\n", dp[k][n]); for (int i = 1; i <= n; i++){ printf("%d", a[i]); printf(" "); if (i == n){ puts(""); break; } if (b[i]) printf("/ "); } } return 0; }