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题目链接:点击打开链接
题意:
给定一个n层书架,一共取m本书。
下面n行给出每层书的价值。
每次可以取任意一层的最左端或最右端的一本书。
问能获得的最大价值。
思路:
1、显然是先求出对于每层任取任意本书能获得的最大价值。
2、然后背包一下。
1:
对于一层书任意j本,那么一定是从左端取k本,右端取 j-k本,求个前缀和然后枚举 j和k即可。每层n^2的dp
2:
分组背包。
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.math.BigInteger; import java.text.DecimalFormat; import java.util.ArrayDeque; import java.util.ArrayList; import java.util.Arrays; import java.util.Collection; import java.util.Collections; import java.util.Comparator; import java.util.Deque; import java.util.HashMap; import java.util.Iterator; import java.util.LinkedList; import java.util.Map; import java.util.PriorityQueue; import java.util.Scanner; import java.util.Stack; import java.util.StringTokenizer; import java.util.TreeMap; import java.util.TreeSet; import java.util.Queue; import java.io.File; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; public class Main { int m, n; int[] sum = new int[N], num = new int[N]; int[][] dp = new int[N][N], h = new int[2][M]; void work() throws Exception{ n = Int(); m = Int(); for(int i = 1; i <= n; i++){ num[i] = Int(); sum[0] = 0; for(int j = 1; j <= num[i]; j++){ sum[j] = sum[j-1]+Int(); } dp[i][0] = 0; for(int j = 1; j <= num[i]; j++){ dp[i][j] = 0; for(int k = 0; k <= j; k++) dp[i][j] = max(dp[i][j], sum[k] + sum[num[i]] - sum[num[i]-j+k]); } } int cur = 0, old = 1; for(int i = 0; i <= m; i++)h[cur][i] = 0; for(int i = 1; i <= n; i++) { cur ^= 1; old ^= 1; for(int j = 0; j <= m; j++)h[cur][j] = h[old][j]; for(int j = 0; j <= num[i]; j++) { for(int k = j; k <= m; k++) h[cur][k] = max(h[cur][k], h[old][k-j]+dp[i][j]); } } out.println(h[cur][m]); } public static void main(String[] args) throws Exception{ Main wo = new Main(); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); // in = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt")))); // out = new PrintWriter(new File("output.txt")); wo.work(); out.close(); } static int N = 101; static int M = 10001; DecimalFormat df=new DecimalFormat("0.0000"); static int inf = (int)1e9; static long inf64 = (long) 1e18; static double eps = 1e-8; static double Pi = Math.PI; static int mod = (int)1e9 + 7 ; private String Next() throws Exception{ while (str == null || !str.hasMoreElements()) str = new StringTokenizer(in.readLine()); return str.nextToken(); } private int Int() throws Exception{ return Integer.parseInt(Next()); } private long Long() throws Exception{ return Long.parseLong(Next()); } private double Double() throws Exception{ return Double.parseDouble(Next()); } StringTokenizer str; static Scanner cin = new Scanner(System.in); static BufferedReader in; static PrintWriter out; /* class Edge{ int from, to, dis, nex; Edge(){} Edge(int from, int to, int dis, int nex){ this.from = from; this.to = to; this.dis = dis; this.nex = nex; } } Edge[] edge = new Edge[M<<1]; int[] head = new int[N]; int edgenum; void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;} void add(int u, int v, int dis){ edge[edgenum] = new Edge(u, v, dis, head[u]); head[u] = edgenum++; }/**/ int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A; int pos = r; r--; while (l <= r) { int mid = (l + r) >> 1; if (A[mid] <= val) { l = mid + 1; } else { pos = mid; r = mid - 1; } } return pos; } int Pow(int x, int y) { int ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } double Pow(double x, int y) { double ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } int Pow_Mod(int x, int y, int mod) { int ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; ans %= mod; y >>= 1; x = x * x; x %= mod; } return ans; } long Pow(long x, long y) { long ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } long Pow_Mod(long x, long y, long mod) { long ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; ans %= mod; y >>= 1; x = x * x; x %= mod; } return ans; } int gcd(int x, int y){ if(x>y){int tmp = x; x = y; y = tmp;} while(x>0){ y %= x; int tmp = x; x = y; y = tmp; } return y; } int max(int x, int y) { return x > y ? x : y; } int min(int x, int y) { return x < y ? x : y; } double max(double x, double y) { return x > y ? x : y; } double min(double x, double y) { return x < y ? x : y; } long max(long x, long y) { return x > y ? x : y; } long min(long x, long y) { return x < y ? x : y; } int abs(int x) { return x > 0 ? x : -x; } double abs(double x) { return x > 0 ? x : -x; } long abs(long x) { return x > 0 ? x : -x; } boolean zero(double x) { return abs(x) < eps; } double sin(double x){return Math.sin(x);} double cos(double x){return Math.cos(x);} double tan(double x){return Math.tan(x);} double sqrt(double x){return Math.sqrt(x);} }
CodeForces 148E Porcelain dp+背包(水
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原文地址:http://blog.csdn.net/qq574857122/article/details/44086723