BZOJ1597: [Usaco2008 Mar]土地购买(dp 斜率优化)

题意

题目链接

sol

重新看了一遍斜率优化，感觉又有了一些新的认识。

首先把土地按照\((w, h)\)排序，用单调栈处理出每个位置第向左第一个比他大的位置，显然这中间的元素是没用的

设\(f[i]\)表示买了前\(i\)块土地的最小花费

\(f[i] = min_{j = 0}^{i - 1}(f[j] + w[i] * h[j + 1])\)

斜率优化一下

// luogu-judger-enable-o2
#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define pt(x) printf("%d ", x);
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;

const int maxn = 1e6 + 10, mod = 1e9 + 7;
ll inf = 6e18 + 10;
const double eps = 1e-9;
template <typename y> inline void chmin(y &a, y b){a = (a < b ? a : b);}
template <typename y> inline void chmax(y &a, y b){a = (a > b ? a : b);}
template <typename y> inline void debug(y a){cout << a << '\n';}
int sqr(int x) {return x * x;}
int add(int x, int y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
void add2(int &x, int y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
int mul(int x, int y) {return 1ll * x * y % mod;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, cur, q[maxn];
ll f[maxn];
struct node {
    int w, h;
    bool operator < (const node &rhs) const {
        return w == rhs.w ? h < rhs.h : w < rhs.w;
    }
}a[maxn];
double y(int x) {
    return f[x];
}
double x(int x) {
    return -a[x + 1].h;
}
double slope(int a, int b) {
    return double(y(b) - y(a)) / (x(b) - x(a));
}
signed main() {
    n = read();
    for(int i = 1; i <= n; i++) a[i].w = read(), a[i].h = read();
    sort(a + 1, a + n + 1);
    for(int i = 1; i <= n; i++) {
        while(cur && a[i].h >= a[cur].h) cur--;
        a[++cur] = a[i]; 
    }
    for(int i = 1; i <= cur; i++) f[i] = inf;
    q[1] = 0;
    for(int i = 1, h = 1, t = 1; i <= cur; i++) {
        while(h < t && slope(q[h], q[h + 1]) < a[i].w) h++;
        f[i] = (ll) f[q[h]] + 1ll * a[i].w * a[q[h] + 1].h;
        while(h < t && slope(q[t], i) < slope(q[t - 1], q[t])) t--; 
        q[++t] = i;
    }
    cout << f[cur];
    return 0;
}