HDU 1300 Pearls(dp)

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题意:

给你一些物品的需求量和单价,你可按照给出的清单吗,也可以将某些珍珠用级别更高的的珍珠来替代,但不能用级别低的替代,让你找到一个购买方案,使得的总价最小。
我已开始想的状态时dp[i][j]表示把第j个里面的一些放在i里面获得的最小值,但是这样的状态太麻烦了,而且是我想多了,因为可以推想要不不拿,要不就整个拿到后面去,
这样就可以保证解的最优性。这样的状态就好办了:dp[i]=min(dp[j]+sum[i]sum[j]p[i])j<i

代码:

////  Created by  CQU_CST_WuErli//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.//#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 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 bignconst 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;const int N=110;int a[N],p[N];int n;int dp[N];int sum[N];int main(int argc, char const *argv[]) {#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    for (int T_T,kase=SI(T_T);kase<=T_T;kase++) {        SI(n);        rep(i,1,n) SII(a[i],p[i]);        sum[0]=0;        rep(i,1,n) sum[i]=i==1?a[i]:sum[i-1]+a[i];        MEM(dp,INF_INT);        dp[0]=0;        rep(i,1,n) rep(j,0,i-1) {            dp[i]=min(dp[i],dp[j]+(sum[i]-sum[j]+10)*p[i]);        }        cout << dp[n] << endl;    }    return 0;}
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