NC50542. 仓库建设
描述
L公司在山上有一些工厂。由于这座山处于高原内陆地区（干燥少雨），L公司一般把产品直接堆放在露天，以节省费用。突然有一天，L公司的总裁L先生接到气象部门的电话，被告知三天之后将有一场暴雨，于是L先生决定紧急在某些工厂建立一些仓库以免产品被淋坏。
输入描述
第一行包含一个整数$N$，表示工厂的个数。
接下来$N$行，每行包含三个整数$X_i,$$P_i,$$C_i$，意义如题中所述。
输出描述
仅包含一个整数，为可以找到最优方案的费用。
示例1
输入:
3 0 5 10 5 3 100 9 6 10
输出:
32
说明:
在工厂1和工厂3建立仓库，建立费用为10+10=20，运输费用为Java(javac 1.8) 解法, 执行用时: 744ms, 内存消耗: 99604K, 提交时间: 2021-03-06 11:14:07
// ps[i] = p[i]*x + ps[i-1];
// s[i] = s[i-1] + p[i];
// dp[i] = dp[j] + c[i] + (s[i] - s[j])*x[i] + (ps[i] - ps[j])
import java.io.*;
import java.lang.reflect.Type;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.*;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.Future;
import java.util.concurrent.FutureTask;
import java.util.concurrent.locks.LockSupport;
public class Main {
static class Task {
public static String roundS(double result, int scale) {
String fmt = String.format("%%.%df", scale);
return String.format(fmt, result);
// DecimalFormat df = new DecimalFormat("0.000000");
// double result = Double.parseDouble(df.format(result));
}
int rt(int x) {
if (x != fa[x]) {
int to = rt(fa[x]);
// dp[x] ^= dp[fa[x]];
fa[x] = to;
return to;
}
return x;
}
void combine(int x, int y, int val) {
int rt1 = rt(x);
int rt2 = rt(y);
if (rt1 == rt2)
return;
fa[rt1] = rt2;
// dp[rt1] = dp[x] ^ dp[y] ^ val;
g--;
}
int fa[];
int g;
static int MAXN = 10000;
static Random rd = new Random(348957438574659L);
static int[] ch[], val, size, rnd, cnt;
static int len = 0, rt = 0;
// return new node, the node below s
static int rotate(int s, int d) {
// child
int x = ch[s][d ^ 1];
// give me grandson
ch[s][d ^ 1] = ch[x][d];
// child become father
ch[x][d] = s;
// update size, update new son first
update(s);
update(x);
return x;
}
static void update(int s) {
size[s] = size[ch[s][0]] + size[ch[s][1]] + cnt[s];
}
// 0 for left, 1 for right
static int cmp(int x, int num) {
if (val[x] == num)
return -1;
return num < val[x] ? 0 : 1;
}
static int insert(int s, int num) {
if (s == 0) {
s = ++len;
val[s] = num;
size[s] = 1;
rnd[s] = rd.nextInt();
cnt[s] = 1;
} else {
int d = cmp(s, num);
if (d != -1) {
ch[s][d] = insert(ch[s][d], num);
// father's random should be greater
if (rnd[s] < rnd[ch[s][d]]) {
s = rotate(s, d ^ 1);
} else {
update(s);
}
} else {
++cnt[s];
++size[s];
}
}
return s;
}
static int del(int s, int num) {
int d = cmp(s, num);
if (d != -1) {
ch[s][d] = del(ch[s][d], num);
update(s);
} else if (ch[s][0] * ch[s][1] == 0) {
if (--cnt[s] == 0) {
s = ch[s][0] + ch[s][1];
}
} else {
int k = rnd[ch[s][0]] < rnd[ch[s][1]] ? 0 : 1;
// k points to smaller random value,then bigger one up
s = rotate(s, k);
// now the node with value num become the child
ch[s][k] = del(ch[s][k], num);
update(s);
}
return s;
}
static int getKth(int s, int k) {
int lz = size[ch[s][0]];
if (k >= lz + 1 && k <= lz + cnt[s]) {
return val[s];
} else if (k <= lz) {
return getKth(ch[s][0], k);
} else {
return getKth(ch[s][1], k - lz - cnt[s]);
}
}
static int getRank(int s, int value) {
if (s == 0)
return 1;
if (value == val[s])
return size[ch[s][0]] + 1;
if (value < val[s])
return getRank(ch[s][0], value);
return getRank(ch[s][1], value) + size[ch[s][0]] + cnt[s];
}
static int getPre(int data) {
int ans = -1;
int p = rt;
while (p > 0) {
if (data > val[p]) {
if (ans == -1 || val[p] > val[ans])
ans = p;
p = ch[p][1];
} else
p = ch[p][0];
}
return ans != -1 ? val[ans] : (-2147483647);
}
static int getNext(int data) {
int ans = -1;
int p = rt;
while (p > 0) {
if (data < val[p]) {
if (ans == -1 || val[p] < val[ans])
ans = p;
p = ch[p][0];
} else
p = ch[p][1];
}
return ans != -1 ? val[ans] : 2147483647;
}
static boolean find(int s, int num) {
while (s != 0) {
int d = cmp(s, num);
if (d == -1)
return true;
else
s = ch[s][d];
}
return false;
}
static int ans = -10000000;
static boolean findX(int s, int num) {
while (s != 0) {
if (val[s] <= num) {
ans = num;
}
int d = cmp(s, num);
if (d == -1)
return true;
else {
s = ch[s][d];
}
}
return false;
}
long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
void linear_sort(int arr[]) {
int d = 65536;
ArrayDeque bucket[] = new ArrayDeque[d];
for (int j = 0; j < d; ++j) {
bucket[j] = new ArrayDeque();
}
for (int u : arr) {
bucket[u % d].offer(u);
}
int pos = 0;
for (int j = 0; j < d; ++j) {
while (bucket[j].size() > 0) {
arr[pos++] = (int)bucket[j].pollFirst();
}
}
for (int u : arr) {
bucket[u / d].offer(u);
}
pos = 0;
for (int j = 0; j < d; ++j) {
while (bucket[j].size() > 0) {
arr[pos++] = (int)bucket[j].pollFirst();
}
}
}
int cur = 0;
int h[];
int to[];
int ne[];
void add(int u, int v) {
to[cur] = v;
ne[cur] = h[u];
h[u] = cur++;
}
public boolean dfs(String cur, HashMap<String, Integer> vis, Map<String, List<String>> list, int cl) {
vis.put(cur, cl);
for (String hp : list.get(cur)) {
if (vis.containsKey(hp)) {
if (vis.get(hp) == cl) {
return false;
}
} else {
if (!dfs(hp, vis, list, 1 - cl)) {
return false;
}
}
}
return true;
}
public class MultiSet {
TreeMap<Integer, Integer> map;
int ct = 0;
MultiSet() {
map = new TreeMap<>();
}
public void remove(int key) {
if (map.containsKey(key)) {
int times = map.get(key);
if (times == 1) {
map.remove(key);
} else {
map.put(key, times - 1);
}
ct--;
}
}
public void add(int key) {
map.put(key, map.getOrDefault(key, 0) + 1);
ct++;
}
public int last() {
return map.lastKey();
}
public int first() {
return map.firstKey();
}
public int size() {
return ct;
}
}
int color[], dfn[], low[], stack[];
int sccno[];
boolean iscut[];
int time = 0, top = 0;
int scc_cnt;
int dcc_cnt;
List<Integer> dcc[];
int root = 0;
// 无向图的强连通分量
void tarjanNonDirect(int u) {
low[u] = dfn[u] = ++time;
stack[top++] = u;
int child = 0;
for (int i = h[u]; i != -1; i = ne[i]) {
int v = to[i];
if (dfn[v] == 0) {
tarjanNonDirect(v);
low[u] = Math.min(low[u], low[v]);
if (low[v] >= dfn[u]) {
if (u != root || ++child > 1) { // 不是root，直接记为cut，是root，判断是否有两个儿子
iscut[u] = true;
}
++dcc_cnt;
// 一个割点可能会被多个点双共享
int z = -1;
do {
z = stack[--top];
//dcc[dcc_cnt].add(z);
} while (z != v);
//dcc[dcc_cnt].add(u);
}
} else {
low[u] = Math.min(low[u], dfn[v]);
// 没有特判是否直接指向父亲
// 回边，使用dfn【v】更新low【u】,因为可能是指向父亲，而父亲的low可能比较小
}
}
}
long dp11[][];
public long s(int l, int r, int a[]) {
if (l > r)
return 0;
if (dp11[l][r] != 0) {
return dp11[l][r];
}
if (l == r) {
return dp11[l][r] = a[l];
}
return dp11[l][r] = Math.max((long)a[l] - s(l + 1, r, a), (long)a[r] - s(l, r - 1, a));
}
int dp[][];
public void solve(int testNumber, InputReader in, PrintWriter out) {
int fk = 1;int n;
int q[] = new int[500010];
while(fk-->0) {
try {
n = in.nextInt();
}catch (Exception e){
break;
}
long x[] = new long[n + 1];
long p[] = new long[n + 1];
long s[] = new long[n + 1];
long c[] = new long[n + 1];
long px[] = new long[n + 1];
for (int i = 1; i <= n; ++i) {
x[i] = in.nextLong();
p[i] = in.nextLong();
s[i] = s[i-1] + p[i];
c[i] = in.nextLong();
px[i] = px[i-1] + p[i]*x[i];
}
long dp[] = new long[n + 1];
int st = 0;
int e = 0;
q[e++] = 0;
for (int i = 1; i <= n; ++i) {
// dp[i] = dp[j] + c[i] + (s[i] - s[j])*x[i] - (px[i] - px[j])
// dp[j] + px[j] = s[j]*x[i] + dp[i] - c[i] + px[i] - x[i]*s[i];
// dp[i] = t[i]*(c[i] - c[0]) + s * (c[n-1] -c[0]);
// for(int j=0;j<i;++j){
// dp[i] = Math.min(dp[i] , dp[j] + c[i]2 + c[j]2 - 2c[i]*c[j] + s))
// dp[j] + c[j]2 = dp[i] -c[i]2 + 2*c[i] * c[j] -s;
// dp[i] = Math.min(dp[i] , dp[j] + (c[i]-c[j])*(t[i]) + s *(c[n-1] - c[j]));
//
// }
// dp[i] = dp[q[st]] + (c[i] - c[q[st]] + s);
//
// while(st<e && dp[q[e-1]] - c[q[e-1]] >= dp[i] - c[i]) e--;
//
// q[e++] = i;
while (st + 1 < e && (dp[q[st + 1]] + px[q[st+1]] - dp[q[st]] - px[q[st]]) <= (s[q[st + 1]] - s[q[st]]) * (x[i])) st++;
dp[i] = dp[q[st]] + c[i] + (s[i] - s[q[st]])*x[i] - (px[i] - px[q[st]]);
while (st + 1 < e && (s[i] - s[q[e - 1]]) * (dp[q[e - 1]] + px[q[e-1]] - dp[q[e - 2]] - px[q[e-2]]) - (dp[i] + px[i] - dp[q[e - 1]] - px[q[e-1]]) * (s[q[e - 1]] - s[q[e - 2]]) >= 0)
e--;
q[e++] = i;
}
out.println(dp[n]);
}
}
static long mul(long a, long b, long p) {
long res = 0, base = a;
while (b > 0) {
if ((b & 1L) > 0)
res = (res + base) % p;
base = (base + base) % p;
b >>= 1;
}
return res;
}
static long mod_pow(long k, long n, long p) {
long res = 1L;
long temp = k % p;
while (n != 0L) {
if ((n & 1L) == 1L) {
res = mul(res, temp, p);
}
temp = mul(temp, temp, p);
n = n >> 1L;
}
return res % p;
}
public static double roundD(double result, int scale) {
BigDecimal bg = new BigDecimal(result).setScale(scale, RoundingMode.UP);
return bg.doubleValue();
}
}
private static void solve() {
InputStream inputStream = System.in;
// InputStream inputStream = null;
// try {
// inputStream = new FileInputStream(new File("D:\\chrome_download\\exp.out"));
// } catch (FileNotFoundException e) {
// e.printStackTrace();
// }
OutputStream outputStream = System.out;
// OutputStream outputStream = null;
// File f = new File("D:\\chrome_download\\");
// try {
// f.createNewFile();
// } catch (IOException e) {
// e.printStackTrace();
// }
// try {
// outputStream = new FileOutputStream(f);
// } catch (FileNotFoundException e) {
// e.printStackTrace();
// }
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task task = new Task();
task.solve(1, in, out);
out.close();
}
public static void main(String[] args) {
new Thread(null, () -> solve(), "1", (1 << 20)).start();
// solve();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String nextLine() {
String line = null;
try {
line = reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
return line;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public char nextChar() {
return next().charAt(0);
}
public int[] nextArray(int n) {
int res[] = new int[n];
for (int i = 0; i < n; ++i) {
res[i] = nextInt();
}
return res;
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}
C++(clang++11) 解法, 执行用时: 359ms, 内存消耗: 24188K, 提交时间: 2020-10-21 14:58:04
#include<bits/stdc++.h>
#define k(i,j) 1.0*(f[j]-f[i])/(p[j]-p[i])
using namespace std;
const int nn=1001021;
int n,t,tt,x[nn],c[nn],q[nn];
long long ans,p[nn],f[nn];
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%lld%d",x+i,p+i,c+i);
for(int i=1;i<=n;i++)
f[0]+=p[i]*(x[n]-x[i]),p[i]+=p[i-1];
ans=(f[0]+=c[n]);
for(int i=1,j;i<n;i++){
if(t>=tt)t=tt;
else while(t<tt&&k(q[t],q[t+1])<x[i]-x[n])t++;
j=q[t],f[i]=f[j]-(p[i]-p[j])*(x[n]-x[i])+c[i];
for(;tt&&k(q[tt-1],q[tt])>=k(q[tt-1],i);tt--);
ans=min(ans,f[q[++tt]=i]);
}return printf("%lld",ans),0;
}