Codeforces 148E Porcelain （dp）

题意：

给定n个序列，每次可以从序列的一个序列的某一段拿一个物品，问最后拿到物品的价值总和最大是多少。略机智的dp，需要预处理出每一个序列拿i个物品能拿到的最大值，然后就变成了一个类似多重背包的东西。我一开始以为是区间DP，发现状态太多不好搞。

代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define rep(flag,start,end) for(int flag=start;flag<=end;flag++)#define Rep(flag,start,end) for(int flag=start;flag>=end;flag--)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bigntemplate <typename T> T gcd(const T& a,const T& b) {return b==0?a:gcd(b,a%b);}const int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;int sum[110][110];int len[110];int n,m;int dp[110][10010];int Max[110][110];int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    while (SII(n,m)==2) {        CLR(sum);        rep(i,1,n) {            SI(len[i]);            rep(j,1,len[i]) SI(sum[i][j]);        }        rep(i,1,n) {            rep(j,2,len[i]) sum[i][j]+=sum[i][j-1];        }        CLR(Max);        rep(i,1,n) {            rep(j,1,len[i]) {                rep(k,0,j) {                    Max[i][j]=max(Max[i][j],sum[i][k]+sum[i][len[i]]-sum[i][len[i]-j+k]);                }            }        }//      rep(i,1,n) {//          rep(j,1,len[i]) cout << Max[i][j] << ' ';//          cout << endl;//      }        int ans=0;        CLR(dp);         for (int i=1;i<=n;i++) {            for (int j=m;j>=0;j--) {                for (int k=0;k<=len[i];k++) {                    if (k<=j) {                        dp[i][j]=max(dp[i][j],dp[i-1][j-k]+Max[i][k]);                    }                }            }        }        cout << dp[n][m] << endl;    }    return 0;}