P2805 [NOI2009]植物大战僵尸 + 最大权闭合子图 X 拓扑排序

传送门：https://www.luogu.org/problemnew/show/P2805

题意

有一个n * m的地图，你可以操纵僵尸从地图的右边向左边走，走的一些地方是有能量值的，有些地方会被一些植物保护起来不能走，只有先吃掉特定植物才能走一些地方。求最大可能拿到的能量值和

思路

最大权闭合子图，由于僵尸只能从一行的右边一步一步走到左边，所以每个格子向右边连一条inf的边（表示选了这个点，右边这个点必选），然后有保护的原因，从被保护的格子向保护的格子连一条inf的边。然后就是最大权闭合子图的操作，源点向正权值格子连容量为这个权值的边，负权值的格子向终点连容量为这个权值绝对值的边。

由于图中有环的存在，我们要把环上以及环之后的点都抹去。

然后跑一遍dinic（），算出最小割，用总的正权值 - 这个最小割就是答案。

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 609;
int val[maxn], vis[maxn], du[maxn];
vector<int> mp[maxn];
vector<pii> ptt[maxn]; //受保护的点。
int n, m;
void topo() {
    queue<int> que;
    for (int i = 1; i <= n * m; i++) {
        if (du[i] == 0) que.push(i);
    }
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        vis[u] = 1;
        for (int i = 0; i < mp[u].size(); i++) {
            int v = mp[u][i];
            du[v]--;
            if (du[v] == 0) que.push(v);
        }
    }
}

struct E {
    int v, w;
    int nxt;
} edge[500009];
int gtot = 0, head[maxn];
void addedge(int u, int v, int w) {
    edge[gtot].v = v;
    edge[gtot].w = w;
    edge[gtot].nxt = head[u];
    head[u] = gtot++;

    edge[gtot].v = u;
    edge[gtot].w = 0;
    edge[gtot].nxt = head[v];
    head[v] = gtot++;
}

int dis[maxn], cur[maxn];
bool bfs(int s, int t) {
    memset(dis, inf, sizeof(dis));
    dis[s] = 0;
    queue<int> que;
    que.push(s);

    for (int i = s; i <= t; i++) cur[i] = head[i];
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v, w = edge[i].w;
            if (w > 0 && dis[v] > dis[u] + 1) {
                dis[v] = dis[u] + 1;
                que.push(v);
            }
        }
    }
    return dis[t] < inf;
}

int dfs(int u, int t, int maxflow) {
    if (u == t || maxflow == 0) return maxflow;
    for (int i = cur[u]; ~i; i = edge[i].nxt) {
        cur[u] = i;
        int v = edge[i].v, w = edge[i].w;
        if (w > 0 && dis[v] == dis[u] + 1) {
            int f = dfs(v, t, min(w, maxflow));
            if (f > 0) {
                edge[i].w -= f;
                edge[i ^ 1].w += f;
                return f;
            }
        }
    }
    return 0;
}
int dinic(int s, int t) {
    int flow = 0;
    while (bfs(s, t)) {
        while (int f = dfs(s, t, inf)) flow += f;
    }
    return flow;
}
int main() {
    memset(head, -1, sizeof(head));
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int u = (i - 1) * m + j, v = u + 1;
            scanf("%d", &val[u]);
            if (j < m) {
                du[u]++;
                mp[v].pb(u); //v -> u
            }
            int q;
            scanf("%d", &q);
            while (q--) {
                int x, y;
                scanf("%d%d", &x, &y);
                x++, y++;
                int p = (x - 1) * m + y;
                du[p]++;
                mp[u].pb(p);
                ptt[u].pb(pii(x, y));
            }
        }
    }

    int s = 0, t = n * m + 1;

    topo();

    int sum = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int u = (i - 1) * m + j, v = u + 1;
            if (vis[u] == 0) continue;
            if (j < m && vis[v]) addedge(u, v, inf);

            if (val[u] >= 0)
                addedge(s, u, val[u]), sum += val[u];
            else
                addedge(u, t, -1 * val[u]);

            for (int k = 0; k < ptt[u].size(); k++) {
                int x = ptt[u][k].fi, y = ptt[u][k].se;
                int p = (x - 1) * m + y;
                addedge(p, u, inf);
            }
        }
    }

    printf("%d\n", sum - dinic(s, t));
    return 0;
}