UESTC - 594 我要长高 单调队列优化DP

【题目链接】 点击打开链接

【题意】中文题目。

【解题方法】 容易想到一个非常朴素的DP。dp[i][j] 代表第i个儿子,身高为j的最低花费。 dp[i][j] = min(dp[i-1][k] + abs(j - k) * C + (x[i] - j) * (x[i] - j));

                      然后分为两种情况:

                     1:第i个儿子的身高,比i-1高时。 dp[i][j] = min(dp[i - 1][k] + j * c - k * c + X), (k <= j) X等于 (x[i] - j) * (x[i] - j);

                     2:第i个儿子的身高,比i-1矮时,dp[i][j] = min(dp[i - 1][k] - j * c + k * c + X), (k >= j) X等于 (x[i] - j) * (x[i] - j);

                     对于第一种情况,我们让f[i - 1][k] = dp[i - 1][k] - k * c;  g[i][j] = j * c + x;


【AC代码】


//
//Created by BLUEBUFF 2016/1/8
//Copyright (c) 2016 BLUEBUFF.All Rights Reserved
//#pragma comment(linker,"/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include  //isstringstream
#include 
#include 
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef pair pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
const int maxn = 220;
const int maxm = 2e5;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9 + 7;
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
//typedef tree,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head
int dp[2][110], n, x, c, q[110], head, tail, cur;
int main()
{while(scanf("%d%d", &n, &c) != EOF){head = tail = cur = 0;CLR(q, 0);scanf("%d", &x);REP2(i, 0, 100){if(i < x) dp[cur][i] = INF;else dp[cur][i] = (x - i) * (x - i);}REP1(i, 1, n){scanf("%d", &x);cur ^= 1;head = tail = 0;REP2(j, 0, 100){int nowf = dp[cur ^ 1][j] - j * c;while(head < tail && q[tail - 1] > nowf) tail--;q[tail++] = nowf;if(j < x) dp[cur][j] = INF;else dp[cur][j] = q[head] + j * c + (j - x) * (j - x);}head = tail = 0;REP3(j, 100, 0){int nowf = dp[cur^1][j] + j * c;while(head < tail && q[tail - 1] > nowf) tail--;q[tail++] = nowf;if(j >= x) dp[cur][j] = min(dp[cur][j], q[head] - j * c + (j - x) * (j - x));else dp[cur][j] = INF;}}int ans = INF;REP2(i, 0, 100){ans = min(ans, dp[cur][i]);}cout << ans << endl;}return 0;
}



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