CodeForces_1348E Phoenix and Berries(动态规划)
Phoenix and Berries
Problem Description
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has ai red berries and bi blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
- the first basket will contain 3 red and 1 blue berries from the first shrub;
- the second basket will contain the 2 remaining red berries from the first
shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1 ≤ n , k ≤ 500 1≤n,k≤500 1≤n,k≤500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a i a_i ai and b i b_i bi ( 0 ≤ a i , b i ≤ 1 0 9 0≤a_i,b_i≤10^9 0≤ai,bi≤109) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Sample Input
2 4
5 2
2 1
Sample Output
2
题意
n棵树，每棵树上有ai个红果实和bi个蓝果实。有可以装k个果实的篮子，一个篮子只能放同种颜色或同一棵树上的果实。求最多可以放满多少个篮子？
题解
显然大多数篮子内的果实都是同颜色的，最多只有n个篮子内的果实是不同色的(若同棵树有多个篮子装的不同色果实，可以将其转为为多个同色篮和一个不同色)。所以只需要考虑每棵树不同颜色的那个篮子的组成。
设 d p [ i ] [ j ] dp[i][j] dp[i][j]为考虑第 i i i棵树果实装完后剩下的红果实数量为 j j j能装满的最大篮子数。
(蓝果实呢？设 d p [ i ] [ j ] [ z ] dp[i][j][z] dp[i][j][z]， z z z代表剩余蓝果实的话，复杂度为 50 0 4 500^4 5004，显然会超时，实际上一个 j j j对应的剩余蓝果实数量是唯一的，等于 果 实 总 数 − d p [ i ] [ j ] ∗ k − j 果实总数-dp[i][j]*k-j 果实总数−dp[i][j]∗k−j)。
状态转移：设第 i i i棵树之前的红蓝果实总数为 s u m sum sum；
枚举不同颜色的那个篮子的组成，由 s 1 s1 s1个红果实和 k − s 1 k-s1 k−s1个蓝果实组成。
则考虑之前的剩余果实，剩余未装篮的红果实有 n u m 1 = j + a [ i ] − s 1 num1 = j+a[i]-s1 num1=j+a[i]−s1个，未装篮的蓝果实数量为 n u m 2 = b [ i ] − ( k − s 1 ) + s u m − d p [ i ] [ j ] ∗ k − j num2 = b[i]-(k-s1)+sum-dp[i][j]*k-j num2=b[i]−(k−s1)+sum−dp[i][j]∗k−j。
所以递推式为
d p [ i + 1 ] [ n u m 1 % k ] = m a x ( d p [ i + 1 ] [ n u m 1 % k ] , d p [ i ] [ j ] + 1 + n u m 1 / k + n u m 2 / k ) dp[i+1][num1\%k] = max(dp[i+1][num1\%k], dp[i][j]+1+num1/k+num2/k) dp[i+1][num1%k]=max(dp[i+1][num1%k],dp[i][j]+1+num1/k+num2/k)
然后考虑不存在不同色篮的转移的情况即可。
取dp[i][j]最大值即为所求。
#include
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 520;
const int mod = 1000000007;
LL dp[maxn][maxn];
int a[maxn], b[maxn];
int main()
{
int n, k, i, j, k2, s1;
LL sum = 0, mx = 0;
memset(dp, -1, sizeof(dp));
scanf("%d %d", &n, &k);
k2 = 2*k;
for(i=0;i<n;i++)
scanf("%d %d", &a[i], &b[i]);
dp[0][0] = 0;
for(i=0;i<n;i++)
{
for(j=0;j<k;j++)
if(dp[i][j]>=0){
int b1 = sum-dp[i][j]*k-j;
for(s1=1;s1<=a[i] && s1<k;s1++)
if(b[i]+s1>=k)
{
int b2 = b1+b[i]-(k-s1);
int a2 = j+a[i]-s1;
dp[i+1][a2%k] = max(dp[i+1][a2%k], dp[i][j]+b2/k+a2/k+1);
}
dp[i+1][(j+a[i])%k] = max(dp[i+1][(j+a[i])%k], dp[i][j]+(j+a[i])/k+(b1+b[i])/k);
}
sum += a[i]+b[i];
}
for(i=0;i<=n;i++)
for(j=0;j<=k;j++)
mx = max(dp[i][j], mx);
printf("%I64d\n", mx);
return 0;
}