[APIO/CTSC2007]数据备份 题解

传送门

首先把数组差分，那么就是要在差分数组中找$k$个不相邻的数，使它们的和最小。

首先有一个暴力的DP做法：$f(i,j,0/1)$表示前$i$个数，选了$j$个，第$i$个数不选/选，和最小是多少。$f(i,j,0)=\min (f(i-1,j,0),f(i-1,j,1)),f(i,j,1)=f(i-1,j-1,0)+d_i$这些转移都不必多说

但是这样复杂度是$O(nk)$的并不能过。解决方法是按套路写个wqs二分优化把$j$那一维降掉就行了。

#include 
#include 
#include 
#include 

template <typename T> inline void read(T& x) {
    int f = 0, c = getchar(); x = 0;
    while (!isdigit(c)) f |= c == '-', c = getchar();
    while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
    if (f) x = -x;
}
template <typename T, typename... Args>
inline void read(T& x, Args&... args) {
    read(x); read(args...); 
}
template <typename T> void write(T x) {
    if (x < 0) x = -x, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + 48);
}
template <typename T> inline void writeln(T x) { write(x); puts(""); }
template <typename T> inline bool chkmin(T& x, const T& y) { return y < x ? (x = y, true) : false; }
template <typename T> inline bool chkmax(T& x, const T& y) { return x < y ? (x = y, true) : false; }

typedef long long LL;
const int maxn = 1e5 + 207;
const LL inf = 1e10;
int n, K;
LL d[maxn];
// f[i][0] = min(f[i-1][0], f[i-1][1])
// f[i][1] = f[i-1][0] + d[i] - mid

struct Data {
    int cnt;
    LL f;
    Data() : cnt(0), f(0) {}
    Data(int a, LL b) : cnt(a), f(b) {}
};
Data dp[maxn][2];
inline bool operator<(const Data &lhs, const Data &rhs) {
    return lhs.f < rhs.f || (lhs.f == rhs.f && lhs.cnt < rhs.cnt);
}

inline void solve(LL mid) {
    for (int i = 0; i <= n; ++i)
        dp[i][0] = dp[i][1] = Data();
    for (int i = 1; i <= n; ++i) {
        dp[i][0] = std::min(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = Data(dp[i - 1][0].cnt + 1, dp[i - 1][0].f + d[i] - mid);
    }
}

int main() {
    read(n, K);
    for (int i = 0; i < n; ++i) read(d[i]);
    for (int i = n - 1; i; --i) d[i] -= d[i - 1];
    d[0] = 0; --n;
    LL left = 0, right = inf, ans;
    while (left <= right) {
        LL mid = (left + right) >> 1;
        solve(mid);
        const Data &mn = std::min(dp[n][0], dp[n][1]);
        if (mn.cnt <= K)
            ans = mn.f + mid * K, left = mid + 1;
        else right = mid - 1;
    }
    writeln(ans);
    return 0;
}