#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 = ;
 int T, n, maxp, w;
 int ap, bp, as, bs;
 int f[maxn][maxn];
 int q[maxn], v[maxn];

 int DP() {
     for (int j = ; j <= maxp; j++)
         f[][j] = -inf;
     for (int i = ; i <= w+; i++) {
         scanf("%d%d%d%d", &ap, &bp, &as, &bs);

         for (int j = ; 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-][j]);
         }
     }
     for (int i = w+; i <= n; i++) {
         scanf("%d%d%d%d", &ap, &bp, &as, &bs);

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

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

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/;
         for (int i = m; i; i--)
             if (__gcd(i, n-i) == ) {
                 printf("%d %d\n", i, n-i);
                 break;
             }
     }
     return ;
 }

854A

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

 int n, k;

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

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

854B

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[];
 long long ans;

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

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

853A

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 = ;
 const int maxday = ;

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

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

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

 int main() {
     cin >> n >> m >> k;
     for (int i = ; 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, );
     for (int i = ; i <= maxday; i++) {
         s[i] = s[i-];
         for (int j = ; j < from[i].size(); j++) {
             if (to[i][j] ==  && 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+);
     for (int i = maxday; i; i--) {
         t[i] = t[i+];
         for (int j = ; j < from[i].size(); j++) {
             if (from[i][j] ==  && 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 = ; i + k <= maxday; i++)
         ans = min(ans, s[i-] + t[i+k]);
     if (ans > 1e12)    puts("-1");
     else    cout << ans << endl;

     return ;
 }

853B

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

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