hdu 3507 Print Article 斜率优化dp

hdu 3507 Print Article

http://acm.hdu.edu.cn/showproblem.php?pid=3507

从http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html这个博客里找的题，讲的很详细，推荐之~~~

dp[i] = MIN(dp[j] + (ans[i]- ans[j])*^2) + m;
若j为最优解，则  dp[j] + (ans[i] - ans[j])^2 + m < dp[k] + (ans[i] - ans[k])^2 + m;
化简：
dp[j] + ans[i]^2 + ans[j]^2 - 2*ans[i]*ans[j] < dp[k] + ans[i]^2 + ans[k]^2 - 2*ans[i]*ans[k];
(dp[j] + ans[j]^2) - (dp[k] + ans[k]^2) < ans[i]*( 2*ans[j] - 2*ans[k]);

若：
(dp[j] + ans[j]^2)  =>  yj
(dp[k] + ans[k]^2)  =>  yk
2*ans[j]  =>  xj
2*ans[k]  =>  xk

则上式为 ( yj - yk ) < a*(xj- xk) 即为斜率

若g[j, k] < a，则j比k更优

若g[j, k] > g[k, l] >a，则k一定不是最优解，可删除，因为g[j, k] > a，则k比j优， 同理，l比k优，故k一定不选

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX(a, b) (a>b?a:b)
#define MIN(a, b) (a<b?a:b)
using namespace std;
const int maxn= 500010;
const int INF= 0x3f3f3f3f;
int dp[maxn], n, m, ans[maxn], que[maxn];
int getx(int j, int k){
    return 2*(ans[j] - ans[k]);
}
int gety(int j, int k){
    return dp[j] - dp[k] + ans[j]*ans[j] - ans[k]*ans[k];
}
void dynamic(){
    int i, j, k, tail, head;
    head= 0, tail= -1;
    que[++tail]= 0;
    for( i= 1; i<= n; i++){
        //找到之前最优的j，得dp[i]值
        while ( head < tail && gety(que[head+1], que[head]) <= ans[i]*getx(que[head+1], que[head]) )
            head++;
        dp[i]= dp[que[head]] + (ans[i] - ans[que[head]])*(ans[i] - ans[que[head]])+ m;
        //找i存放的位置
        while ( head < tail && gety(i, que[tail])*getx(que[tail], que[tail-1]) <= gety(que[tail], que[tail-1])*getx(i, que[tail]))
            tail--;
        que[++tail]= i;
    }
}
int main(){
    //freopen("1.txt", "r", stdin);
    int i, j, k, aa;
    while( scanf("%d%d", &n, &m) != EOF){
        ans[0]= 0;
        for( i= 1; i<= n; i++){
            scanf("%d", &aa);
            ans[i]= ans[i-1] + aa;
        }
        dp[0]= 0;
        dynamic();
        printf("%d\n", dp[n]);
    }
    return 0;
}