四边形优化dp

理解：

http://blog.renren.com/share/263498909/1064362501

http://www.cnblogs.com/ronaflx/archive/2011/03/30/1999764.html

http://yomean.blog.163.com/blog/static/189420225201272864127683/

http://www.cnblogs.com/zxndgv/archive/2011/08/02/2125242.html

题目总结：

http://www.cnblogs.com/ronaflx/archive/2011/03/30/1999764.html

下摘自：http://www.cnblogs.com/zxndgv/archive/2011/08/02/2125242.html

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最有代价用d[i,j]表示 

d[i,j]=min{d[i,k-1]+d[k+1,j]}+w[i,j] 

当中w[i,j]=sum[i,j] 

四边形不等式   

     w[a,c]+w[b,d]<=w[b,c]+w[a,d](a<b<c<d) 就称其满足凸四边形不等式 

决策单调性 

     w[i,j]<=w[i',j']   ([i,j]属于[i',j']) 既 i'<=i<j<=j'



于是有下面三个定理 



定理一： 假设w同一时候满足四边形不等式 和 决策单调性 ,则d也满足四边形不等式

定理二：当定理一的条件满足时，让d[i,j]取最小值的k为K[i,j]，则K[i,j-1]<=K[i,j]<=K[i+1,j] 

定理三：w为凸当且仅当w[i,j]+w[i+1,j+1]<=w[i+1,j]+w[i,j+1] 



由定理三知 推断w是否为凸即推断 w[i,j+1]-w[i,j]的值随着i的添加是否递减 

于是求K值的时候K[i,j]仅仅和K[i+1,j] 和 K[i,j-1]有关。所以 能够以i-j递增为顺序递推各个状态值终于求得结果  将O(n^3)转为O(n^2)

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注意：

注意决策单调性顺序，既要符合决策调性，也要符合题意 (!!!!)

注意枚举顺序，依据dp方程和决策单调性方程

注意初始化，防止訪问到无效状态或没处理的状态。

dp和s边界初始化，尤其是决策的上届和下届初始化。

例题1：

石子合并问题：hdu 3506

dp[i][j] = min{dp[i][k] + dp[k + 1][j] + cost[i, j] }, i <= k <= j - 1 , cost[i][j] = sum[j] - sum[i - 1]

s[i][j - 1] <= s[i][j] <= s[i + 1][j]

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FD(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)

typedef long long LL;
const int INF = 1000000007;
const double eps = 1e-10;
const int maxn = 2010;

int dp[maxn][maxn], s[maxn][maxn];
int w[maxn][maxn];

int n, m;
int val[maxn];
int sum[maxn];

///求dp最小值
///枚举 区间 由小到大
void solve()
{
//    memset(dp, 0, sizeof(dp));///初始化无效值
    FE(i, 1, 2 * n)
    {
        dp[i][i] = 0;
        s[i][i] = i;/// 初始化决策下届,为0
    }
    FE(len, 2, n)
    {
        for (int i = 2 * n - len; i >= 1; i--)
        {
            int j = i + len - 1;

            dp[i][j] = INF;
            int a = s[i][j - 1], b = s[i + 1][j];
            int cost = sum[j] - sum[i - 1];
            for (int k = a; k <= b; k++)
            {
                if (dp[i][j] > dp[i][k] + dp[k + 1][j] + cost)
                {
                    dp[i][j] = dp[i][k] + dp[k + 1][j] + cost;
                    s[i][j] = k;
                }
            }
        }
    }
}

int main ()
{
    while (~RI(n))
    {
        FE(i, 1, n) RI(val[i]), val[i + n] = val[i];;
        FE(i, 1, 2 * n) sum[i] = sum[i - 1] + val[i];
//        pre();
        solve();
        int ans = INF;
        FE(i, 1, n)
        {
            if (ans > dp[i][n + i - 1])
                ans = dp[i][n + i - 1];
        }
        printf("%d\n", ans);
    }
    return 0;
}

例题2：邮局问题：

id=1160">poj 1160

1:注意此法的 i 和 j 顺序与寻常不同

此时决策区间为：s[i - 1][j] <= s[i][j] <= s[i][j + 1] (!!!)

详细见凝视:

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FD(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)

typedef long long LL;
const int INF = 1000000007;
const double eps = 1e-10;
const int maxn= 1000010;

int dp[33][333], s[33][333];
int w[333][333];

int n, m;
int val[333];
int sum[333];
///dp[i][j] = min{dp[i - 1][k] + w[k + 1][j]}, i - 1 <= k <= j - 1
///一般要求 i <= j （！！）
///s[i - 1][j] <= s[i][j] <= s[i][j + 1] （！

！

）
///注意决策单调性顺序，既要符合决策调性，也要符合题意 (!!!!)
///注意枚举顺序，依据dp方程和决策单调性方程
///注意初始化，防止訪问到无效状态或没处理的状态。dp和s边界初始化。尤其是决策的上届和下届初始化。

void pre()
{
    for(int i = 1; i <= n; i ++) //这里有一个递推公式能够进行预处理
    {
        w[i][i] = 0;
        for(int j = i + 1; j <= n; j ++)
        {
            int mid = (j + i) >> 1;
            w[i][j] = w[i][j - 1] + val[j] - val[mid];
//            int x = sum[j] - sum[mid] - val[mid] * (j - mid);
//            x += val[mid] * (mid - i) - (sum[mid - 1] - sum[i - 1]);
        }
    }
}

///求dp最小值
///枚举i从小到大
///再枚举j从大到小
void solve()
{
    memset(dp, 0, sizeof(dp));///初始化无效值
    FE(i, 1, n)
    {
        dp[1][i] = w[1][i];
        s[1][i] = 0;/// 初始化决策下届,为0
    }
    FE(i, 2, m)
    {
        //s[1][i] = 0;
        s[i][n + 1] = n;///初始化决策上届
        for (int j = n; j >= i; j--)
        {
            int tmp = dp[i][j] = INF;///初始化最优值
            int a = s[i - 1][j], b = s[i][j + 1];
            //a = max(a, i - 1); b = min(b, j - 1); // i - 1 <= k <= j - 1
            for (int k = a; k <= b; k++) ///保证枚举到的都是有效状态，且都已计算过
            {
                if (tmp > dp[i - 1][k] + w[k + 1][j])
                {
                    tmp = dp[i - 1][k] + w[k + 1][j];
                    s[i][j] = k;
                }
            }
            dp[i][j] = tmp;
        }
    }
}

int main ()
{
    while (~RII(n, m))
    {
        FE(i, 1, n) RI(val[i]), sum[i] = sum[i - 1] + val[i];
        pre();
        solve();
        printf("%d\n", dp[m][n]);
    }
    return 0;
}

2:-详细见凝视

此时决策区间为：s[i][j - 1] <= s[i][j] <= s[i + 1][j] (!!!)

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FD(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)

typedef long long LL;
const int INF = 1000000007;
const double eps = 1e-10;
const int maxn= 1000010;

int dp[333][33], s[333][33];
int w[333][333];

int n, m;
int val[333];
int sum[333];
///dp[i][j] = min{dp[k][j - 1] + w[k + 1][j]}, j - 1 <= k <= i - 1
///此处i>=j
///s[i][j - 1] <= s[i][j] <= s[i + 1][j] （！！

）
///注意决策单调性顺序，既要符合决策调性，也要符合题意 (!!!!)
///注意枚举顺序。依据dp方程和决策单调性方程
///注意初始化。防止訪问到无效状态或没处理的状态。

dp和s边界初始化，尤其是决策的上届和下届初始化。

void pre()
{
    for(int i = 1; i <= n; i ++) //这里有一个递推公式能够进行预处理
    {
        w[i][i] = 0;
        for(int j = i + 1; j <= n; j ++)
        {
            int mid = (j + i) >> 1;
            w[i][j] = w[i][j - 1] + val[j] - val[mid];
        }
    }
}

///求dp最小值
///枚举i从小到大
///再枚举j从大到小
void solve()
{
    memset(dp, 0, sizeof(dp));///初始化无效值
    FE(i, 1, n)
    {
        dp[i][1] = w[1][i];
        s[i][1] = 0;/// 初始化决策下届,为0
    }
    FE(i, 2, m)
    {
        //s[i][1] = 0;
        s[n + 1][i] = n;///初始化决策上届
        for (int j = n; j >= i; j--)
        {
            int tmp = dp[j][i] = INF;///初始化最优值
            int a = s[j][i - 1], b = s[j + 1][i];
            //a = max(a, i - 1); b = min(b, j - 1);
            for (int k = a; k <= b; k++) ///保证枚举到的都是有效状态，且都已计算过
            {
                if (tmp > dp[k][i - 1] + w[k + 1][j])
                {
                    tmp = dp[k][i - 1] + w[k + 1][j];
                    s[j][i] = k;
                }
            }
            dp[j][i] = tmp;
        }
    }
}

int main ()
{
    while (~RII(n, m))
    {
        FE(i, 1, n) RI(val[i]);
        pre();
        solve();
        printf("%d\n", dp[n][m]);
    }
    return 0;
}