bzoj 1283 序列 - 费用流

题目传送门

　　传送门

题目大意

　　给定一个长度为$n$的序列，要求选出一些数使得原序列中每$m$个连续的数中不超过$K$个被选走。问最大的可能的和。

　　感觉建图好妙啊。。

　　考虑把问题转化成选$m$次数，每次选出一个子序列， 要求两次选择的数的下标差至少为$m$。

　　这个很容易建图就能做。$i$向$\min\{T, i + m\}$连边，容量为1，费用为$a_i$。$i$向$i + 1$连边容量为1，费用为0.

Code

 /**
  * bzoj
  * Problem#1283
  * Accepted
  * Time: 232ms
  * Memory: 1436k
  */
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <queue>
 using namespace std;
 typedef bool boolean;

 template <typename T>
 void pfill(T* pst, const T* ped, T val) {
     for ( ; pst != ped; *(pst++) = val);
 }

 typedef class Edge {
     public:
         int ed, nx, r, c;

         Edge(int ed = , int nx = , int r = , int c = ) : ed(ed), nx(nx), r(r), c(c) {    }
 } Edge;

 typedef class MapManager {
     public:
         int* h;
         vector<Edge> es;

         MapManager() {    }
         MapManager(int n) {
             h = new int[(n + )];
             pfill(h, h + n + , -);
         }

         void addEdge(int u, int v, int r, int c) {
             es.push_back(Edge(v, h[u], r, c));
             h[u] = (signed) es.size() - ;
         }

         void addArc(int u, int v, int cap, int c) {
             addEdge(u, v, cap, c);
             addEdge(v, u, , -c);
         }

         Edge& operator [] (int p) {
             return es[p];
         }
 } MapManager;

 const signed int inf = (signed) (~0u >> );

 typedef class Graph {
     public:
         int S, T;
         MapManager g;

         int *le;
         int *f, *mf;
         boolean *vis;

         Graph() {    }
         // be sure T is the last node
         Graph(int S, int T) {
             this->S = S;
             this->T = T;
             f = new int[(T + )];
             le = new int[(T + )];
             mf = new int[(T + )];
             vis = new boolean[(T + )];
             pfill(vis, vis + T, false);
         }

         int spfa() {
             queue<int> que;
             pfill(f, f + T + , -inf);
             que.push(S);
             f[S] = , le[S] = -, mf[S] = inf;
             while (!que.empty()) {
                 int e = que.front();
                 que.pop();
                 vis[e] = false;
                 for (int i = g.h[e], eu, w; ~i; i = g[i].nx) {
                     if (!g[i].r)
                         continue;
                     eu = g[i].ed, w = f[e] + g[i].c;
                     if (w > f[eu]) {
                         f[eu] = w, le[eu] = i, mf[eu] = min(mf[e], g[i].r);
                         if (!vis[eu]) {
                             vis[eu] = true;
                             que.push(eu);
                         }
                     }
                 }
             }
             if (f[T] == -inf)
                 return inf;
             int rt = ;
             for (int p = T, e; ~le[p]; p = g[e ^ ].ed) {
                 e = le[p];
                 g[e].r -= mf[T];
                 g[e ^ ].r += mf[T];
                 rt += mf[T] * g[e].c;
             }
             return rt;
         }

         int max_cost() {
             int rt = , delta;
             while ((delta = spfa()) != inf) {
                 rt += delta;
 //                cerr << delta << '\n';
             }
             return rt;
         }
 } Graph;

 int n, m, k;

 inline void solve() {
     scanf("%d%d%d", &n, &m, &k);
     Graph graph(, n + );
     MapManager& g = graph.g;
     k = min(k, m);
     g = MapManager(n + );
     g.addArc(, , k, );
     for (int i = , x; i <= n; i++) {
         scanf("%d", &x);
         g.addArc(i, min(n + , i + m), , x);
         g.addArc(i, i + , k, );
     }
     printf("%d\n", graph.max_cost());
 }

 int main() {
     solve();
     return ;
 }