题解

直接看大佬的题解吧 · 题解 P2805 【[NOI2009]植物大战僵尸】

最大权 = 正点权和 - 最小割

---

---

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
bool vis[N];
int n, m, k;
int score[N];
int id(int x,int y){return x*m+y;}

namespace Network_flows { //网络流板子
    //设定起点和终点
    int st;//起点-源点
    int ed;//终点-汇点

    struct egde {
        int to, next;
        int flow;//剩余流量
    } e[N * 2];

    int head[N], tot = 1;

    void add(int u, int v, int w) {
        e[++tot] = {v, head[u], w};
        head[u] = tot;
        e[++tot] = {u, head[v], 0};
        head[v] = tot;//网络流反相边流量为0
    }

    int dep[N];//dep[]=-1时为炸点
    queue<int> q;

    bool bfs() {
        memset(dep, 0, sizeof(dep));//顺便起到vis的功能
        q.push(st);
        dep[st] = 1;
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int i = head[u]; i; i = e[i].next) {
                int v = e[i].to;
                if (!dep[v] && e[i].flow) {
                    dep[v] = dep[u] + 1;
                    q.push(v);
                }
            }
        }
        return dep[ed];
    }

    int dfs(int u, int Flow) {
        if (u == ed) return Flow;
        int now_flow = 0;//跑残流
        for (int i = head[u]; i; i = e[i].next) {
            int v = e[i].to;
            if (dep[v] == dep[u] + 1 && e[i].flow) {
                int f = dfs(v, min(Flow - now_flow, e[i].flow));
                e[i].flow -= f;
                e[i ^ 1].flow += f;
                now_flow += f;
                if (now_flow == Flow) return Flow;
            }
        }
        if (now_flow == 0)dep[u] = -1;
        return now_flow;
    }

	#define max_flow dinic
    int dinic() {//最大流
        int res = 0;
        while (bfs()) {
            res += dfs(st, INF);
        }
        return res;
    }

    void init() {
        tot = 1;
        memset(head, 0, sizeof(head));
        while (!q.empty()) q.pop();
    }
}
using namespace Network_flows;


vector<int> out[N];
int in[N];//入度

void topo(){
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if(!in[id(i,j)]){
                q.push(id(i,j));
                vis[id(i,j)]=true;//可以攻击到的植物
            }
        }
    }
    
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for (int i = 0; i <out[u].size(); ++i) {
            int v=out[u][i];
            in[v]--;
            if(!vis[v] && !in[v]){
                q.push(v);
                vis[v]=true;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(0);

    cin>>n>>m;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int cnt;
            cin>>score[id(i,j)]>>cnt;
            for (int l = 1,x,y; l <= cnt; ++l) {
                cin>>x>>y;
                x++;y++;//编号从0开始
                out[id(i,j)].push_back(id(x,y));//负责保护的指向被保护的
                in[id(x,y)]++;
            }
            
            if(j<m){
                out[id(i,j+1)].push_back(id(i,j));//右边的保护左边的
                in[id(i,j)]++;
            }
        }
    }

    topo();//找到进攻顺序 排除相互保护永远无法被吃掉的植物
    init();//清空队列 初始化网络流
    
    st=0,ed=id(n+1,m+1);
    int sum=0;//统计所有正权之和
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int u=id(i,j);
            if(!vis[u])continue;//被保护着无法吃掉
            if(score[u]>=0){//点权正的和源点连
                add(st,u,score[u]);
                sum+=score[u];
            }else {//点权负的和汇点连
                add(u,ed,-score[u]);
            }

            for (int l = 0; l <out[u].size(); ++l) {
                int v=out[u][l];
                if(vis[u]){//被保护的连向负责保护 边权设置为INF表示无法被删除
                    add(v,u,INF);
                }
            }
        }
    }
    cout <<sum -dinic()<< endl; //最大权 = 正点权和 - 最小割 
    return 0;
}

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来源： https://blog.csdn.net/Yubing792289314/article/details/104501207