BZOJ 4873 寿司餐厅 网络流

 

最大权闭合子图

1.每个区间收益(i,j)对应一个点 权值为正连S 负连T

2.每个区间收益向其子区间收益(i+1,j)与(i,j-1)对应的两个点连边 容量为INF

3.每个寿司类型对应一个点 连一条边到T 容量为m*w[i]*w[i]

4.每个寿司对应的区间收益点(i,i)连一条边到对应的寿司类型 容量为INF 再连一条边到T 容量为w[i]

最后跑最大流

//Netflow dumpling
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 11111;
const int MAXM = 100000;
const int INF = 1000000050;
int Head[MAXN], cur[MAXN], lev[MAXN], to[MAXM << 1], nxt[MAXM << 1], f[MAXM << 1], ed = 1, S, T;
inline void addedge(int u, int v, int cap)
{
    to[++ed] = v;
    nxt[ed] = Head[u];
    Head[u] = ed;
    f[ed] = cap;
    to[++ed] = u;
    nxt[ed] = Head[v];
    Head[v] = ed;
    f[ed] = 0;
    return;
}
inline bool BFS()
{
    int u;
    memset(lev, -1, sizeof(lev));
    queue<int>q;
    lev[S] = 0;
    q.push(S);
    while (q.size()) {
        u = q.front();
        q.pop();
        for (int i = Head[u]; i; i = nxt[i])
            if (f[i] && lev[to[i]] == -1) {
                lev[to[i]] = lev[u] + 1;
                q.push(to[i]);
                /*
                if (to[i] == T)
                {
                        return 1;
                }
                magic one way optimize
                */
            }
    }
    memcpy(cur, Head, sizeof Head);
    return lev[T] != -1;
}
inline int DFS(int u, int maxf)
{
    if (u == T || !maxf) {
        return maxf;
    }
    int cnt = 0;
    for (int &i = cur[u], tem; i; i = nxt[i])
        if (f[i] && lev[to[i]] == lev[u] + 1) {
            tem = DFS(to[i], min(maxf, f[i]));
            maxf -= tem;
            f[i] -= tem;
            f[i ^ 1] += tem;
            cnt += tem;
            if (!maxf) {
                break;
            }
        }
    if (!cnt) {
        lev[u] = -1;
    }
    return cnt;
}
int Dinic()
{
    int ans = 0;
    while (BFS()) {
        ans += DFS(S, 2147483647);
    }
    return ans;
}
void init(int SS, int TT)
{
    memset(Head, 0, sizeof(Head));
    ed = 1;
    S = SS;
    T = TT;
    return;
}
int n, m, now;
bool ok[1005];
int a[105];
int getid(int x, int y)
{
    return 1000 + n * (x - 1) + y;
}
int main()
{
    int ans = 0;
    scanf("%d %d", &n, &m);
    S = 0, T = n * n + 1005;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (!ok[a[i]]) {
            addedge(a[i], T, m * a[i]*a[i]), ok[a[i]] = 1;
        }
        addedge(getid(i, i), a[i], INF);
        addedge(getid(i, i), T, a[i]);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            scanf("%d", &now);
            if (now > 0) {
                addedge(S, getid(i, j), now);
                ans += now;
            } else {
                addedge(getid(i, j), T, -now);
            }
            if (i != j) {
                addedge(getid(i, j), getid(i + 1, j), INF);
                addedge(getid(i, j), getid(i, j - 1), INF);
            }
        }
    }
    ans -= Dinic();
    printf("%d\n", ans);
    return 0;
}