【HDU6611】费用流

HDU6611
分析：
这题竟然不是dp
题意是求k个不下降子序列的和的最大值。
竟然不是dp！！！

将i拆点，容量1，费用-a[i]，
然后对于a[j]>=a[i],连i+n到j，容量1，费用0
然后起点连i,i+n连终点。跑容量为k的最小费用即可。
最后答案取反。

正确性很显然。
把图画出来就可以完全理解了。
//电子科大板子真好用

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int maxn = 1e4;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double Pi = acos(-1.0);
inline int read_int() {
	char c;
	int ret = 0, sgn = 1;
	do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
	if (c == '-') sgn = -1; else ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
	return sgn * ret;
}
inline ll read_ll() {
	char c;
	ll ret = 0, sgn = 1;
	do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
	if (c == '-') sgn = -1; else ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
	return sgn * ret;
}
struct edge {
	int to, capacity, cost, rev;
	edge() {}
	edge(int to, int _capacity, int _cost, int _rev) :to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
struct Min_Cost_Max_Flow {
	int V, H[maxn + 5], dis[maxn + 5], PreV[maxn + 5], PreE[maxn + 5];
	vector<edge> G[maxn + 5];
	//调用前初始化
	void Init(int n) {
		V = n;
		for (int i = 0; i <= V; ++i)G[i].clear();
	}
	//加边
	void Add_Edge(int from, int to, int cap, int cost) {
		G[from].push_back(edge(to, cap, cost, G[to].size()));
		G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
	}
	//flow是自己传进去的变量，就是最后的最大流，返回的是最小费用
	int Min_cost_max_flow(int s, int t, int f, int& flow) {
		int res = 0; fill(H, H + 1 + V, 0);
		while (f) {
			priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
			fill(dis, dis + 1 + V, INF);
			dis[s] = 0; q.push(pair<int, int>(0, s));
			while (!q.empty()) {
				pair<int, int> now = q.top(); q.pop();
				int v = now.second;
				if (dis[v] < now.first)continue;
				for (int i = 0; i < G[v].size(); ++i) {
					edge& e = G[v][i];
					if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {
						dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
						PreV[e.to] = v;
						PreE[e.to] = i;
						q.push(pair<int, int>(dis[e.to], e.to));
					}
				}
			}
			if (dis[t] == INF)break;
			for (int i = 0; i <= V; ++i)H[i] += dis[i];
			int d = f;
			for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].capacity);
			f -= d; flow += d; res += d*H[t];
			for (int v = t; v != s; v = PreV[v]) {
				edge& e = G[PreV[v]][PreE[v]];
				e.capacity -= d;
				G[v][e.rev].capacity += d;
			}
		}
		return res;
	}
	int Max_cost_max_flow(int s, int t, int f, int& flow) {
		int res = 0;
		fill(H, H + 1 + V, 0);
		while (f) {
			priority_queue <pair<int, int> > q;
			fill(dis, dis + 1 + V, -INF);
			dis[s] = 0;
			q.push(pair<int, int>(0, s));
			while (!q.empty()) {
				pair<int, int> now = q.top(); q.pop();
				int v = now.second;
				if (dis[v] > now.first)continue;
				for (int i = 0; i < G[v].size(); ++i) {
					edge& e = G[v][i];
					if (e.capacity > 0 && dis[e.to] < dis[v] + e.cost + H[v] - H[e.to]) {
						dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
						PreV[e.to] = v;
						PreE[e.to] = i;
						q.push(pair<int, int>(dis[e.to], e.to));
					}
				}
			}
			if (dis[t] == -INF)break;
			for (int i = 0; i <= V; ++i)H[i] += dis[i];
			int d = f;
			for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].capacity);
			f -= d; flow += d;
			res += d*H[t];
			for (int v = t; v != s; v = PreV[v]) {
				edge& e = G[PreV[v]][PreE[v]];
				e.capacity -= d;
				G[v][e.rev].capacity += d;
			}
		}
		return res;
	}
};
int a[maxn],n,k;
Min_Cost_Max_Flow MCMF;
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&k);
        MCMF.Init(2*n+10);
        int s=0,t=2*n+5;
        //MCMF.Add_Edge(s,s2,k,0);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            MCMF.Add_Edge(i,i+n,1,-a[i]);
        }
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                if(a[j]>=a[i]){
                    MCMF.Add_Edge(i+n,j,INF,0);
                }
            }
        }
        for(int i=1;i<=n;i++){
            MCMF.Add_Edge(s,i,INF,0);
            MCMF.Add_Edge(i+n,t,INF,0);
        }
        int flow0;
        printf("%d\n",-MCMF.Min_cost_max_flow(s,t,k,flow0));
    }
    return 0;
}