#10：wannanewtry——6

HDU3401，列完转移方程拆分一下，正着、反着跑优先队列优化代表买或卖。初始化不大会搞……

#include <bits/stdc++.h>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 2005;
int T, n, maxp, w;
int ap, bp, as, bs;
int f[maxn][maxn];
int q[maxn], v[maxn];

int DP() {
    for (int j = 1; j <= maxp; j++)
        f[0][j] = -inf;
    for (int i = 1; i <= w+1; i++) {
        scanf("%d%d%d%d", &ap, &bp, &as, &bs);
        
        for (int j = 0; j <= maxp; j++) {
            if (j <= as)    f[i][j] = -j*ap;
            else    f[i][j] = -inf;
            f[i][j] = max(f[i][j], f[i-1][j]);
        }
    }
    for (int i = w+2; i <= n; i++) {
        scanf("%d%d%d%d", &ap, &bp, &as, &bs);
        
        int l = 1, r = 0;
        for (int j = 0; j <= maxp; j++) {
            f[i][j] = f[i-1][j];
            while (l <= r && f[i-w-1][j] + j*ap > v[r])    r--;
            q[++r] = j;
            v[r] = f[i-w-1][j] + j*ap;
            
            while (l <= r && q[l] < j-as)    l++;
            if (l <= r)    f[i][j] = max(f[i][j], v[l]-j*ap);
        }

        l = 1, r = 0;
        for (int j = maxp; ~j; j--) {
            while (l <= r && f[i-w-1][j] + j*bp > v[r])    r--;
            q[++r] = j;
            v[r] = f[i-w-1][j] + j*bp;

            while (l <= r && q[l] > j+bs)    l++;
            if (l <= r)    f[i][j] = max(f[i][j], v[l]-j*bp);
        }
    }
    return f[n][0];
}

int main() {
    for (cin >> T; T; T--) {
        cin >> n >> maxp >> w;
        cout << DP() << endl;
    }
    return 0;
}

为了少一些痛苦自闭，蒟蒻只好写几道cf水题。

#include <bits/stdc++.h>
using namespace std;

int n;

int main() {
    while (cin >> n) {
        int m = n/2;
        for (int i = m; i; i--)
            if (__gcd(i, n-i) == 1) {
                printf("%d %d
", i, n-i);
                break;
            }
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

int n, k;

int main() {
    while (cin >> n >> k) {
        if (k > 0 && k < n)
            printf("1 ");
        else    printf("0 ");

        printf("%d
", min(n - k, 2*k));
    }
    return 0;
}

优先队列乱模拟一下：

#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;
int n, k;
priority_queue<P> Q;
set<int> minute;
set<int>::iterator it;
int v[300005];
long long ans;

int main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        int cost;
        scanf("%d", &cost);
        Q.push(P(cost, i));
        minute.insert(i+k);
    }

    for (int i = 1; i <= n; i++) {
        auto t = Q.top(); Q.pop();
        int k1 = t.second;
        if (k1 < k)    k1 = k;
        it = minute.lower_bound(k1);
        v[t.second] = *it;
        ans += (long long)(*it - t.second) * t.first;
        minute.erase(it);
    }

    cout << ans << endl;
    for (int i = 1; i <= n; i++)
        printf("%d ", v[i]);
    return 0;
}

差分预处理，枚举区间：

#include <bits/stdc++.h>
#define ll long long
#define ri readint()
#define gc getchar()
#define pb push_back
#define P pair<int, int>
using namespace std;

const ll INF = 1e13;
const int maxn = 100005;
const int maxday = 1000000;

int n, m, k;
ll f[maxn];
ll s[maxday+5], t[maxday+5];
vector<int> from[maxday+5], to[maxday+5], cost[maxday+5];

inline int readint() {
    int x = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x*10 + c - 48;
    return x * s;
}

void init(ll *x, int k) {
    for (int i = 1; i <= n; i++)
        f[i] = INF;
    x[k] = INF*n;
}

int main() {
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
        int day = ri, f = ri, t = ri, c = ri;
        from[day].pb(f);
        to[day].pb(t);
        cost[day].pb(c);
    }

    init(s, 0);
    for (int i = 1; i <= maxday; i++) {
        s[i] = s[i-1];
        for (int j = 0; j < from[i].size(); j++) {
            if (to[i][j] == 0 && cost[i][j] < f[from[i][j]]) {
                s[i] -= f[from[i][j]] - cost[i][j];
                f[from[i][j]] = cost[i][j];
            }
        }
    }
    init(t, maxday+1);
    for (int i = maxday; i; i--) {
        t[i] = t[i+1];
        for (int j = 0; j < from[i].size(); j++) {
            if (from[i][j] == 0 && cost[i][j] < f[to[i][j]]) {
                t[i] -= f[to[i][j]] - cost[i][j];
                f[to[i][j]] = cost[i][j];
            }
        }
    }

    ll ans = INF;
    for (int i = 2; i + k <= maxday; i++)
        ans = min(ans, s[i-1] + t[i+k]);
    if (ans > 1e12)    puts("-1");
    else    cout << ans << endl;

    return 0;
}

BZOJ1010，斜率优化dp经典题，推式子很重要，设定不同的函数做法也会有难易区分。

#include <bits/stdc++.h>
#define db double
#define maxn 50005
using namespace std;

int n, L;
int head, tail;
int q[maxn];
db sum[maxn], f[maxn];

db sqr(db x) {
    return x * x;
}
db a(int i) {//斜率单调的
    return sum[i] + i;
}
db b(int i) {
    return sum[i] + i + L + 1;
}
db Y(int i) {
    return f[i] + sqr(b(i));
}
db slope(int i, int j) {
    return (Y(j) - Y(i)) / (b(j) - b(i));
}

int main() {
    cin >> n >> L;
    for (int i = 1; i <= n; i++) {
        scanf("%lf", &sum[i]);
        sum[i] += sum[i-1];
    }
    head = 1, tail = 1;
    for (int i = 1; i <= n; i++) {
        while (head < tail && slope(q[head], q[head+1]) < 2*a(i))    head++;
        f[i] = f[q[head]] + sqr(a(i) - b(q[head]));

        while (head < tail && slope(q[tail-1], q[tail]) > slope(q[tail-1], i))    tail--;
        q[++tail] = i;
    }
    cout << fixed << setprecision(0) << f[n] << endl;
    return 0;
}