Kattis peaktram (树状数组 DP)

题目链接:点击这里

题意:给出n个物体的高度以及将他们高度改变1单位的花费,一个观测点从最左端从0往上走,一直平视. 问至少能观测到k个物体的最小花费.

因为高度比较大,所以需要将高度离散化,那么需要离散化哪些高度呢?容易发现对于每一个建筑,只要把它+-70范围内的高度离散化就好了,这样最多就是70*140种高度.然后用dp[i][j][k]表示到i物体,最大高度是j,一共看到k个物体的最小花费,转移式子就是

dp[i][j][k]=min{d.p[i−1][j′][k−1]+cost(j,hi)dp[i−1][j][k]+(j≤hi?cost(hi,j):0)(j′<j)(j′≥j)

不难发现对于下面一半可以直接转移,对于上面一半可以维护一个树状数组,总复杂度大约是O(n4lgn).

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <stack>
#define Clear(x,y) memset (x,y,sizeof(x))
#define Close() ios::sync_with_stdio(0)
#define Open() freopen ("more.in", "r", stdin)
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<long long, int>
#define pb push_back
#define mod 1000000007
template <class T>
inline bool scan (T &ret) {
char c;
int sgn;
if (c = getchar(), c == EOF) return 0; //EOF
while (c != '-' && (c < '0' || c > '9') ) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
using namespace std;
#define maxn 75
#define maxm maxn*maxn*2
#define INF 0x3f3f3f3f3f3f3f3f

long long dp[maxn][maxm][maxn];
long long h[maxn], cost[maxn];
vector <int> num;
long long c[maxn][maxn][maxm];
int n, len, sz;

int lowbit (int x) {
return x&(-x);
}

void update (int x, long long num, long long *c) {
for (int i = x; i < maxm; i += lowbit (i)) {
c[i] = min (c[i], num);
}
}

long long query (int x, long long *c) {
long long ans = INF;
for (int i = x; i > 0; i -= lowbit (i)) {
ans = min (ans, c[i]);
}
return ans;
}

void solve () {
sz = num.size ();
for (int i = 1; i <= sz; i++) {
dp[1][i][1] = abs (h[1]-num[i-1])*cost[1];
update (i, dp[1][i][1], c[1][1]);
}
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= sz; j++) {
for (int k = 1; k <= n; k++) {
dp[i][j][k] = min (dp[i][j][k], dp[i-1][j][k] + (num[j-1]<=h[i] ? cost[i]*abs(num[j-1]-h[i]) : 0));
long long tmp = query (j-1, c[i-1][k-1]);
dp[i][j][k] = min (dp[i][j][k],
tmp+cost[i]*abs (h[i]-num[j-1]));
update (j, dp[i][j][k], c[i][k]);
}
}
}
long long ans = INF;
for (int i = 1; i <= sz; i++) {
for (int j = len; j <= n; j++) {
ans = min (ans, dp
[i][j]);
}
}
cout << ans << endl;
}

int main () {
//Open ();
Close ();
num.clear ();
Clear (c, INF);
Clear (dp, INF);
cin >> n >> len;
for (int i = 1; i <= n; i++) {
cin >> h[i] >> cost[i];
for (int j = 0; j <= 70; j++) {
num.pb (h[i]+j);
}
for (int j = 1; j <= 70 && h[i]-j > 0; j++) {
num.pb (h[i]-j);
}
}
sort (num.begin (), num.end ());
num.erase (unique (num.begin (), num.end ()), num.end ());
solve ();
return 0;
}