CF321E Ciel and Gondolas 【决策单调性dp】

题目链接

CF321E

题解

题意：将\(n\)个人分成\(K\)段，每段的人两两之间产生代价，求最小代价和
容易设\(f[k][i]\)表示前\(i\)个人分成\(k\)段的最小代价和
设\(val(i,j)\)为\(i\)到\(j\)两两之间产生的代价和，容易发现就是一个矩形，可以预处理前缀和\(O(1)\)计算
那么有
\[f[k][i] = min\{f[k - 1][j] + val(j + 1,i)\}\]
直接转移显然\(O(n^2k)\)
我们把\(val(j + 1,i)\)拆开，也不能分离\(i\)和\(j\)
很好，可以大胆猜想这个\(dp\)是符合决策单调性的

证明：
如果对于\(x > y\)且\(x\)作为\(i\)的决策，一定有
\[f[k - 1][x] + val(x + 1,i) \le f[k - 1][y] + val(y + 1,i)\]
那么对于\(i' > i\)，由几何面积可以得知\(val(x + 1,i') - val(x + 1,i) \le val(y + 1,i') - val(y + 1,i)\)
所以仍有
\[f[k - 1][x] + val(x + 1,i') \le f[k - 1][y] + val(y + 1,i')\]
故对于\(i\)决策时比\(y\)更优的\(x\)，在\(i' > i\)的决策时依旧更优
即该\(dp\)满足决策单调性
证毕

所以用一个队列维护三元组，即可做到\(O(n^2 + nk)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 4005,maxm = 805,INF = 0x3f3f3f3f;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
struct tri{int l,r,pos;}q[maxn];
int head,tail;
int s[maxn][maxn],f[maxm][maxn],n,K,now;
inline int val(int i,int j){
    return s[i][i] - s[i][j - 1] - s[j - 1][i] + s[j - 1][j - 1];
}
inline bool check(int pos,int i,int j){
    return f[now - 1][i] + val(pos,i + 1) <= f[now - 1][j] + val(pos,j + 1);
}
inline void work(){
    f[now][0] = INF;
    q[head = tail = 0] = (tri){1,n,0};
    tri u;
    for (register int i = 1; i <= n; i++){
        u = q[head];
        f[now][i] = f[now - 1][u.pos] + val(i,u.pos + 1);
        q[head].l++;
        if (q[head].l > q[head].r) head++;
        while (head <= tail){
            u = q[tail--];
            if (check(u.l,i,u.pos)){
                if (head > tail) {q[++tail] = (tri){u.l,n,i}; break;}
                continue;
            }
            else if (!check(u.r,i,u.pos)){
                q[++tail] = u;
                if (u.r == n) break;
                q[++tail] = (tri){u.r + 1,n,i};
                break;
            }
            else {
                int l = i + 1,r = n,mid;
                while (l < r){
                    mid = l + r >> 1;
                    if (check(mid,i,u.pos)) r = mid;
                    else l = mid + 1;
                }
                q[++tail] = (tri){u.l,l - 1,u.pos};
                q[++tail] = (tri){l,n,i};
                break;
            }
        }
    }
}
int main(){
    n = read(); K = read();
    REP(i,n) REP(j,n) s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + read();
    REP(i,n) f[1][i] = val(i,1); f[1][0] = INF;
    for (now = 2; now <= K; now++) work();
    printf("%d\n",f[K][n] >> 1);
    return 0;
}