nefu-dp1 （线性dp）
nefu-dp1
https://vjudge.csgrandeur.cn/contest/571200#overview
感谢z神的题单
dp废物来打基础了。
(感觉难度大概是递减的)
琪露诺
单调队列优化dp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int f[N], a[N], n, l, r; //f[i]:i结尾的最大值
int main () {
cin >> n >> l >> r;
for (int i = 0; i <= n; i++) cin >> a[i];
queue<int> q; //单调队列优化dp
memset (f, -0x3f, sizeof f);
f[0] = a[0];
q.push (0);
for (int i = l; i <= n; i++) {
while (!q.empty () && f[q.front ()] < f[i-l]) q.pop ();
q.push (i - l);
while (!q.empty() && q.front () < i - r) q.pop ();
f[i] = f[q.front ()] + a[i];
}
int ans = f[n];
for (int i = 1; i < r; i++) ans = max (ans, f[n-i]);
cout << ans << endl;
}
护卫队
要用到一个dp辅助数组，提前预处理！
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1005;
const double inf = 1e18;
ll n, s, W, w[N];
double f[N], t[N][N], tt[N]; //f[i]: 前i辆车通过的最短时间, t[i][j]: i-j位一车队的通过时间
int main () {
cin >> W >> s >> n;
for (int i = 1; i <= n; i++) {
f[i] = inf;
for (int j = 1; j <= n; j++) {
t[i][j] = inf;
}
}
for (int i = 1; i <= n; i++) {
int x;
cin >> w[i] >> x;
tt[i] = 1.0 * s / x * 60;
t[i][i] = tt[i];
w[i] += w[i-1];
}
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
t[i][j] = max (t[i][j-1], tt[j]);
//cout << t[i][j] << ' ';
}
//cout << endl;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++) {
if (w[i] - w[j-1] > W) continue;
f[i] = min (f[i], f[j-1] + t[j][i]);
}
}
cout << fixed << setprecision (1) << f[n];
}
书的复制
重点在于方案输出，pre数组记录
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 505;
int n, m, pre[N];
ll a[N], s[N], f[N][N]; //f[i][j]: 考虑到第i人,划分了j组的最小抄写时间
int main () {
memset (f, 0x3f, sizeof f);
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i], s[i] = s[i-1] + a[i];
for (int i = 1; i <= n; i++) f[i][1] = s[i];
for (int i = 1; i <= n; i++) {
int id = 0;
for (int j = 1; j <= min (i, m); j++) {
for (int k = 1; k < i; k++) {
f[i][j] = min (f[i][j], max (f[k][j-1], s[i] - s[k]));
}
//cout << f[i][j] << ' ';
}
//cout << endl;
}
//cout << f[n][m] << endl;
int s = 0, w = n;
for(int i = n; i >= 1; i--){//这里贪心，因为后面的人抄尽量多，前面少，所以倒过来枚举
s += a[i];
if (s > f[n][m]) s = a[i], pre[w] = i + 1, w = i;//pre[w]记录这个人抄书的开始位置
}
cout << 1 << ' ' << w << endl;//第一组特判
//output (n);
//for (int i = 1; i <= n; i++) cout << pre[i] << ' ';
for (int i = 1; i <= n; i++) {
if (pre[i]) cout << pre[i] << ' ' << i << endl;
}
}
摆花
重点在方案涉及，二维即可
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 105, mod = 1e6 + 7;
int n, m, a[N];
ll f[N][N]; //f[i][j]: 前i种花盆摆了j个的方案数
int main () {
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i];
f[0][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= m; j++) { //该种花也可以一个都不摆
for (int k = 0; k <= min(j, a[i]); k++) {
(f[i][j] += f[i-1][j-k]) %= mod;
}
}
}
cout << f[n][m] << endl;
}
导弹拦截
二分优化求LIS
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int a[N], f[N]; //f[i]: 长度为i的结尾最大值
int main () {
int n = 0, x, len;
while (cin >> x) a[++n] = x;
f[1] = a[1], len = 1;
for (int i = 2; i <= n; i++) {
if (a[i] <= f[len]) f[++len] = a[i];
else {
int l = 1, r = len;
while (l < r) {
int mid = l + r >> 1;
if (f[mid] < a[i]) r = mid;
else l = mid + 1;
}
f[r] = a[i];
}
}
//for (int i = 1; i <= len; i++) cout << f[i] << ' ';
cout << len << endl;
//不升序列个数等于最长上升序列长度
f[1] = a[1], len = 1;
for (int i = 1; i <= n; i++) {
if (a[i] > f[len]) f[++len] = a[i];
else {
int l = 1, r = len;
while (l < r) {
int mid = l + r >> 1;
if (f[mid] >= a[i]) r = mid;
else l = mid + 1;
}
f[l] = a[i];
}
}
//for (int i = 1; i <= len; i++) cout << f[i] << ' ';
cout << len << endl;
}
开心的金明
萌萌题
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 3e4 + 5, M = 30, mod = 1e6 + 7;
int n, m;
ll f[M][N], ans; //f[i][j]: 前i个物品剩j元的最大价值
int main () {
cin >> m >> n;
for (int i = 1; i <= n; i++) {
int a, b;
cin >> a >> b;
for (int j = m; j >= 0; j--) {
f[i][j] = f[i-1][j];
if (j - a >= 0) f[i][j] = max (f[i-1][j], f[i-1][j-a] + 1ll * a * b);
}
}
for (int i = 0; i <= m; i++) ans = max (ans, f[n][i]);
cout << ans << endl;
}
小A点菜
萌萌题
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 105, M = 10005;
int n, m, a;
ll f[N][M]; //f[i][j]: 前i个菜剩j元的方案
int main () {
cin >> n >> m;
f[0][0] = 1;
for (int i = 1; i <= n; i++) {
cin >> a;
for (int j = m; j >= 0; j--) {
f[i][j] += f[i-1][j];
if (j >= a) f[i][j] += f[i-1][j-a];
}
}
cout << f[n][m] << endl;
}
砝码称重
萌萌题
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 10005;
int a[] = {1, 2, 3, 5, 10, 20}, ans, x;
bool f[N];
int main () {
f[0] = 1;
for (int i = 0; i < 6; i++) {
cin >> x;
for (int j = 1000; j >= 0; j--) {
for (int k = x; k > 0; k--) {
f[j + k * a[i]] |= f[j];
}
}
}
// for (int i = 0; i <= 1000; i++) {
// if (f[i]) cout << i << endl;
// }
for (int i = 1; i <= 1000; i++) ans += f[i];
cout << "Total=" << ans << endl;
}
