HDU 4106 Fruit Ninja interval k coverage problem minimum cost flow

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Test instructions

Given n-long sequences, M, K

Select some numbers to make the selected number and maximum. Output and.

Limit: The maximum number of k in any interval [I, I+m] is selected.

Ideas:

Petition P367, Interval K coverage problem, an interval as a point, then select a point is equivalent to covering the M range.

#include <iostream> #include <stdio.h> #include <string.h> #include <queue> #include <math.h >using namespace std; #define LL int#define inf 0x3f3f3f3f#define inf 0x3fffffffffffffffll#define N 3000#define M 3000*3    000struct Edge {ll to, cap, cost, NEX; Edge () {} edge (ll to,ll cap,ll cost,ll next): To, Cap (CAP), cost, NEX (next) {}} edge[m];ll Head[n], Edgenum;ll D[n]    , A[n], P[n];bool inq[n];void Add (ll from,ll to,ll cap,ll cost) {Edge[edgenum] = Edge (To,cap,cost,head[from]);    Head[from] = edgenum++;    Edge[edgenum] = Edge (From,0,-cost,head[to]); Head[to] = edgenum++;}    BOOL SPFA (ll S, ll T, LL &flow, ll &cost) {for (ll i = 0; I <= t; i++) d[i] = inf;    memset (inq, 0, sizeof inq);    queue<ll>q;    Q.push (s); D[s] = 0;    A[s] = inf;        while (!q.empty ()) {ll u = Q.front (); Q.pop ();        Inq[u] = 0;            for (ll i = head[u]; ~i; i = Edge[i].nex) {Edge &e = Edge[i]; if (E.cap && D[e.to] > D[u] + e.cost) {d[e.to] = D[u] + e.cost;                P[e.to] = i;                A[e.to] = min (A[u], e.cap);            if (!inq[e.to]) {inq[e.to]=1; Q.push (e.to);}    }}}//if the cost is INF, abort the charge flow if (d[t] = = inf) return false;    Cost + = d[t] * A[t];    Flow + = A[t];    ll u = t;        while (U! = s) {edge[p[u]].cap-= a[t];        Edge[p[u]^1].cap + = a[t];    u = edge[p[u]^1].to; } return true;    ll Mincost (ll S,ll t) {ll flow = 0, cost = 0;    while (SPFA (s, t, flow, cost)); return cost;} void Init () {memset (head,-1,sizeof head); edgenum = 0;}    int A[n], from, to, N, M, k;void input () {for (int i = 1; i<= N; i++) scanf ("%d", &a[i]);    Init (); from = 0;    to = n+2;    Add (from, 1, K, 0);        for (int i = 1; I <= n; i++) {int tmp = MIN (m+i, n+1);        Add (i, TMP, 1,-a[i]);    Add (i, i+1, K, 0); } Add (N+1, to, K, 0);} int main () {while (~SCANF ("%d%d%d", &n, &amP;m, &k)) {input ();        int cost = Mincost (from, to);    cout<<-cost<<endl; } return 0;}

HDU 4106 Fruit Ninja interval k coverage problem minimum cost flow