bzoj 1283 序列

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 = 0, int nx = 0, int r = 0, int c = 0) : 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 + 1)];
            pfill(h, h + n + 1, -1);
        }

        void addEdge(int u, int v, int r, int c) {
            es.push_back(Edge(v, h[u], r, c));
            h[u] = (signed) es.size() - 1;
        }

        void addArc(int u, int v, int cap, int c) {
            addEdge(u, v, cap, c);
            addEdge(v, u, 0, -c);
        }

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

const signed int inf = (signed) (~0u >> 1);

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 + 1)];
            le = new int[(T + 1)];
            mf = new int[(T + 1)];
            vis = new boolean[(T + 1)];
            pfill(vis, vis + T, false);
        }
        
        int spfa() {
            queue<int> que;
            pfill(f, f + T + 1, -inf);
            que.push(S);
            f[S] = 0, le[S] = -1, 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 = 0;
            for (int p = T, e; ~le[p]; p = g[e ^ 1].ed) {
                e = le[p];
                g[e].r -= mf[T];
                g[e ^ 1].r += mf[T];
                rt += mf[T] * g[e].c;
            }
            return rt;
        }

        int max_cost() {
            int rt = 0, delta;
            while ((delta = spfa()) != inf) {
                rt += delta;
//                cerr << delta << '
';
            }
            return rt;
        }
} Graph;

int n, m, k;

inline void solve() {
    scanf("%d%d%d", &n, &m, &k);
    Graph graph(0, n + 1);
    MapManager& g = graph.g;
    k = min(k, m);
    g = MapManager(n + 1);
    g.addArc(0, 1, k, 0);
    for (int i = 1, x; i <= n; i++) {
        scanf("%d", &x);
        g.addArc(i, min(n + 1, i + m), 1, x);
        g.addArc(i, i + 1, k, 0);
    }
    printf("%d
", graph.max_cost());
}

int main() {
    solve();
    return 0;
}