二进制分组优化多重背包
index = 0;
for (int i = 1; i <= m; i++) {
int c = 1, p, h, k;
cin >> p >> h >> k;
while (k > c) {
k -= c;
list[++index].w = c * p;
list[index].v = c * h;
c *= 2;
}
list[++index].w = p * k;
list[index].v = h * k;
}
单调队列优化多重背包
#include <array>
#include <deque>
#include <iostream>
constexpr int MAXV = 4e4 + 10;
constexpr int MAXN = 1e2 + 10;
using namespace std;
int n, W, last = 0, now = 1;
array<int, MAXN> v, w, k;
array<array<int, MAXV>, 2> f;
deque<int> q;
int main() {
ios::sync_with_stdio(false);
cin >> n >> W;
for (int i = 1; i <= n; i++) {
cin >> v[i] >> w[i] >> k[i];
}
for (int i = 1; i <= n; i++) {
for (int y = 0; y < w[i]; y++) {
// 清空队列
deque<int>().swap(q);
for (int x = 0; x * w[i] + y <= W; x++) {
// 弹出不在范围的元素
while (!q.empty() && q.front() < x - k[i]) {
q.pop_front();
}
// 保证队列单调
while (!q.empty() && f[last][q.back() * w[i] + y] - q.back() * v[i] <
f[last][x * w[i] + y] - x * v[i]) {
q.pop_back();
}
q.push_back(x);
f[now][x * w[i] + y] =
f[last][q.front() * w[i] + y] - q.front() * v[i] + x * v[i];
}
}
swap(last, now);
}
cout << f[last][W] << endl;
return 0;
}
数位dp
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
ull f[20][10],power[20];
void init() {
power[0]=1;
for(int i=1;i<=19;i++) power[i]=power[i-1]*10;
for(int i=0;i<=9;i++) f[1][i]=1;
for(int i=2;i<=19;i++)
for(int j=0;j<=9;j++)
for(int k=0;k<=9;k++)
if(abs(j-k)>=2)
f[i][j]+=f[i-1][k];
return ;
}
ull solve(ull x) {
int w=0,y,pre;
ull ans=0;
while(power[w]<=x) w++;
for(int i=1;i<w;i++)
for(int j=1;j<=9;j++)
ans+=f[i][j];
y=x/power[w-1];
for(int j=1;j<y;j++) ans+=f[w][j];
pre=y,x%=power[w-1];
for(int i=w-1;i>=1;i--) {
y=x/power[i-1];
for(int j=0;j<y;j++)
if(abs(j-pre)>=2)
ans+=f[i][j];
if(abs(pre-y)<2) break;
pre=y,x%=power[i-1];
}
return ans;
}
int main() {
ull a,b;
cin>>a>>b;
init();
cout<<solve(b+1)-solve(a)<<endl;
return 0;
}
斜率优化（玩具装箱）
#include<bits/stdc++.h>
using namespace std;
#define db double
#define int long long
const int maxn = 1e6+10;
int n,L,v[maxn];
int f[maxn],sum[maxn];
int q[maxn],hd,tl;
inline int read(){
int x = 0,f = 0;
char c = getchar();
while(!isdigit(c))
f|=(c=='-'),c = getchar();
while(isdigit(c))
x = x*10+c-'0',c = getchar();
if(f)x = -x;
return x;
}
db a(int x){
return sum[x]+x;
}
db b(int x){
return sum[x]+x+L+1;
}
db X(int x){
return a(x);
}
db Y(int x){
return f[x]+b(x)*b(x);
}
db slope(int x,int y){
return (Y(y) - Y(x))/(X(y) - X(x));
}
signed main(){
#ifndef ONLINE_JUDGE
freopen("in.in","r",stdin);
freopen("out.out","w",stdout);
#endif
n = read(),L = read();
for(int i=1;i<=n;i++){
int x = read();
sum[i] = sum[i-1]+x;
}
hd = tl = 1;
for(int i=1;i<=n;i++){
while(hd<tl&&slope(q[hd],q[hd+1])<2*a(i))
hd++;
f[i] = f[q[hd]]+(a(i) - b(q[hd]))*(a(i) - b(q[hd]));
while(hd<tl&&slope(q[tl-1],q[tl])>slope(q[tl-1],i))
tl--;
q[++tl] = i;
}
printf("%lld\n",f[n]);
return 0;
}
二分队列四边形不等式（诗人小G）
#include <bits/stdc++.h>
#define ll long double
using namespace std;
const int maxn = 1e6 + 10;
int T;
int p, n, pr[maxn], nxt[maxn], h, t;
ll a[maxn], s[maxn], le, f[maxn], q[maxn], r[maxn];
char c[maxn][40];
ll qpow(ll a, int b) {
ll res = 1;
while (b) {
if (b & 1)
res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
ll calc(int i, int j) {
return (f[j] + qpow(abs(s[i] - s[j] - 1 - le), p));
}
int bound(int x, int y) {
int l = x, r = n + 1;
while (l < r) {
int mid = (l + r) >> 1;
if (calc(mid, x) >= calc(mid, y))
r = mid;
else
l = mid + 1;
}
return l;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
freopen("out.out", "w", stdout);
#endif
scanf("%d", &T);
while (T--) {
scanf("%d%Lf%d", &n, &le, &p);
memset(f, 0x3f, sizeof(f));
memset(pr, 0, sizeof(pr));
memset(nxt, 0, sizeof(nxt));
for (int i = 1; i <= n; ++i) {
scanf("%s", c[i]);
a[i] = (ll)strlen(c[i]) + 1;
s[i] = s[i - 1] + a[i];
}
q[h = t = 1] = 0, r[h = t = 1] = n;
f[0] = 0;
for (int i = 1; i <= n; ++i) {
while (h < t && r[h] < i)
++h;
f[i] = calc(i, q[h]), pr[i] = q[h];
while (h < t && r[t - 1] >= bound(q[t], i))
--t;
r[t] = bound(q[t], i) - 1, q[++t] = i;
}
if (f[n] > 1e18) {
puts("Too hard to arrange");
}
else {
printf("%0.Lf\n", f[n]);
int x = n;
while (pr[x]) {
nxt[pr[x]] = x;
x = pr[x];
}
nxt[0] = x;
x = 0;
while (nxt[x]) {
for (int i = x + 1; i < nxt[x]; ++i) {
printf("%s ", c[i]);
}
printf("%s\n", c[nxt[x]]);
x = nxt[x];
}
}
puts("--------------------");
}
return 0;
}
分治决策单调性（魔法商店）
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 2e6 + 10;
namespace IO {
void openfile() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
freopen("out.out", "w", stdout);
#endif
}
inline int read() {
int x = 0, f = 0;
char c = getchar();
while (!isdigit(c))
f |= c == '-', c = getchar();
while (isdigit(c))
x = x * 10 + c - '0', c = getchar();
if (f)
x = -x;
return x;
}
} // namespace IO
using namespace IO;
int n, m, k;
int sum[10][maxn];
int c[maxn], v[maxn], sta[10][maxn], top[maxn];
int f[10][maxn], now[maxn], num;
void dp(int pos, int l, int r, int x, int y) {
if (l > r)
return;
int mid = (l + r) >> 1, p = mid;
for (int i = max(x, mid - top[pos]); i <= min(mid, y); i++)
if (f[pos - 1][now[i]] + sum[pos][mid - i] >= f[pos - 1][now[p]] + sum[pos][mid - p])
p = i;
//这个位置一定要是>=因为p如果是mid可能会大于等于y，这样右区间就没有决策区间了
f[pos][now[mid]] = f[pos - 1][now[p]] + sum[pos][mid - p];
dp(pos, l, mid - 1, x, p), dp(pos, mid + 1, r, p, y);
}
bool cmp(int x, int y) {
return x > y;
}
signed main() {
openfile();
n = read(), m = read(), k = read();
for (int i = 1; i <= n; i++) {
c[i] = read(), v[i] = read();
sta[c[i]][++top[c[i]]] = v[i];
}
for (int i = 1; i <= 7; i++)
sort(sta[i] + 1, sta[i] + 1 + top[i], cmp);
for (int i = 1; i <= 7; i++)
for (int j = 1; j <= top[i]; j++)
sum[i][j] = sum[i][j - 1] + sta[i][j];
memset(f, -0x7f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= 7; i++) {
for (int j = 0; j < min(m + 1, i); j++) {
for (int k = j; k <= m; k += i)
now[++num] = k;
dp(i, 1, num, 1, num);
for (int k = 1; k <= num; k++)
now[k] = 0;
num = 0;
}
}
// for (int i = 1; i <= 7; i++) {
// for (int j = 0; j <= m; j++)
// cout << f[i][j] << ' ';
// cout << endl;
// }
int ans = 0;
for (int i = 1; i <= m; i++) {
ans = max(ans, f[7][i]);
printf("%lld\n", ans);
}
cerr << 1.0 * clock() / CLOCKS_PER_SEC << '\n';
return 0;
}
slope trick（P4331 [BalticOI 2004] Sequence 数字序列）
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pii pair<int, int>
#define mp make_pair
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;
namespace IO {
void openfile() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
freopen("out.out", "w", stdout);
#endif
}
int mul(int x, int y) {
return 1ll * x * y % mod;
}
int add(int x, int y) {
x += y;
if (x >= mod)
x -= mod;
return x;
}
int sub(int x, int y) {
return add(x, mod - y);
}
inline int read() {
int x = 0, f = 0;
char c = getchar();
while (!isdigit(c))
f |= c == '-', c = getchar();
while (isdigit(c))
x = x * 10 + c - '0', c = getchar();
if (f)
x = -x;
return x;
}
} // namespace IO
using namespace IO;
int n, a[maxn];
int top, ans;
int b[maxn], ls[maxn], rs[maxn], tot, dis[maxn];
struct node {
int rt, l, r, siz, val;
} sta[maxn];
int merge(int x, int y) {
if (!x || !y)
return x + y;
if (a[x] < a[y])
swap(x, y);
rs[x] = merge(rs[x], y);
if (dis[ls[x]] < dis[rs[x]])
swap(ls[x], rs[x]);
dis[x] = dis[rs[x]] + 1;
return x;
}
int del(int x) {
return merge(ls[x], rs[x]);
}
signed main() {
openfile();
dis[0] = -1;
n = read();
for (int i = 1; i <= n; i++)
a[i] = read() - i;
for (int i = 1; i <= n; i++) {
sta[++top] = {i, i, i, 1, a[i]};
while (top != 1 && sta[top - 1].val > sta[top].val) {
top--;
sta[top].rt = merge(sta[top].rt, sta[top + 1].rt);
sta[top].siz += sta[top + 1].siz;
sta[top].r = sta[top + 1].r;
while (sta[top].siz > (sta[top].r - sta[top].l + 2) / 2)
sta[top].siz--, sta[top].rt = del(sta[top].rt);
sta[top].val = a[sta[top].rt];
}
}
int cnt = 1;
for (int i = 1; i <= n; i++) {
if (sta[cnt].r < i)
++cnt;
ans += abs(sta[cnt].val - a[i]);
}
printf("%lld\n", ans);
cnt = 1;
for (int i = 1; i <= n; i++) {
if (sta[cnt].r < i)
++cnt;
printf("%lld ", sta[cnt].val + i);
}
puts("");
cerr << 1.0 * clock() / CLOCKS_PER_SEC << '\n';
return 0;
}
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