BZOJ 1565 Luogu P2805 [NOI2009]植物大战僵尸 (Tarjan判环、最小割)

我: “立个flag 14点之前调完这题”

洛谷AC时间: 2019-06-24 14:00:16

实力打脸。。。

网络流板子从来写不对系列

题目链接: (BZOJ) https://www.lydsy.com/JudgeOnline/problem.php?id=1565

(luogu) https://www.luogu.org/problemnew/show/P2805

题解: 长得就那么像个最大权闭合子图啊。。。

\(i\)攻击\(j\)相当于如果想吃掉\(i\)必须吃掉\(j\), 另外如果吃掉\(i\)必须吃掉\(i\)右侧的点，最大权闭合子图。

但是可能有环，注意环并非要么全吃要么全不吃而是全都不能吃，所以用Tarjan提前判一下环，在环里的点和\(T\)连\(+\inf\)表示不能吃

时间复杂度\(O(MaxFlow(nm,(nm)^2))\).

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 602;
const int M = 362400;
const int INF = 1e8;

namespace MaxFlow
{
    struct Edge
    {
        int v,w,nxt,rev;
    } e[(M<<1)+3];
    int fe[N+3];
    int te[N+3];
    int dep[N+3];
    int que[N+3];
    int n,en;
    void addedge(int u,int v,int w)
    {
        en++; e[en].v = v; e[en].w = w;
        e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;
        en++; e[en].v = u; e[en].w = 0;
        e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
    }
    bool bfs()
    {
        for(int i=1; i<=n; i++) dep[i] = 0;
        int head = 1,tail = 1; que[tail] = 1; dep[1] = 1;
        while(head<=tail)
        {
            int u = que[head]; head++;
            for(int i=fe[u]; i; i=e[i].nxt)
            {
                if(dep[e[i].v]==0 && e[i].w>0)
                {
                    dep[e[i].v] = dep[u]+1;
                    tail++; que[tail] = e[i].v;
                }
            }
        }
        return dep[2]!=0;
    }
    int dfs(int u,int cur)
    {
        if(u==2) {return cur;}
        int rst = cur;
        for(int i=te[u]; i; i=e[i].nxt)
        {
            if(dep[e[i].v]==dep[u]+1 && e[i].w>0 && rst>0)
            {
                int flow = dfs(e[i].v,min(rst,e[i].w));
                if(flow>0)
                {
                    e[i].w-=flow; e[e[i].rev].w += flow; rst-=flow;
                    if(e[i].w>0) {te[u] = i;}
                    if(rst==0) {return cur;}
                }
            }
        }
        if(cur==rst) {dep[u] = 0;}
        return cur-rst;
    }
    int dinic(int _n)
    {
        n = _n;
        int ret = 0;
        while(bfs())
        {
            for(int i=1; i<=n; i++) te[i] = fe[i];
            ret += dfs(1,INF);
        }
        return ret;
    }
}

namespace Tarjan
{
    struct Edge
    {
        int v,nxt;
    } e[M+3];
    int fe[N+3];
    int dfn[N+3],low[N+3],stk[N+3];
    bool instk[N+3];
    bool inscc[N+3];
    int n,en,cnt,tp;
    void addedge(int u,int v)
    {
        en++; e[en].v = v;
        e[en].nxt = fe[u]; fe[u] = en;
    }
    void dfs(int u)
    {
        cnt++; dfn[u] = low[u] = cnt; tp++; stk[tp] = u; instk[u] = true;
        for(int i=fe[u]; i; i=e[i].nxt)
        {
            if(!dfn[e[i].v])
            {
                dfs(e[i].v);
                low[u] = min(low[u],low[e[i].v]);
            }
            else if(instk[e[i].v])
            {
                low[u] = min(low[u],low[e[i].v]);
            }
        }
        if(low[u]>=dfn[u])
        {
            if(stk[tp]!=u) {inscc[u] = true;}
            while(stk[tp]!=u)
            {
                inscc[stk[tp]] = true;
                instk[stk[tp]] = false;
                stk[tp] = 0; tp--;
            }
            stk[tp] = 0; tp--; instk[u] = false;
        }
    }
    void work(int _n)
    {
        n = _n;
        for(int i=1; i<=n; i++)
        {
            if(!dfn[i]) {dfs(i);}
        }
    }
}

int n,m;

int getid(int x,int y) {return x*m+y+3;}

int main()
{
    scanf("%d%d",&n,&m); int ans = 0;
    for(int i=1; i<=n*m; i++)
    {
        int x; scanf("%d",&x);
        if(x>0) {MaxFlow::addedge(1,i+2,x); ans += x;}
        else if(x<0) {MaxFlow::addedge(i+2,2,-x);}
        scanf("%d",&x);
        while(x--)
        {
            int y,z; scanf("%d%d",&y,&z); int pos = getid(y,z);
            MaxFlow::addedge(pos,i+2,INF);
            Tarjan::addedge(pos-2,i);
        }
        if(i%m!=0)
        {
            MaxFlow::addedge(i+2,i+3,INF);
            Tarjan::addedge(i,i+1);
        }
    }
    Tarjan::work(n*m);
    for(int i=1; i<=n*m; i++)
    {
        if(Tarjan::inscc[i])
        {
            MaxFlow::addedge(i+2,2,INF);
        }
    }
    int tmp = MaxFlow::dinic(n*m+2);
    ans -= tmp;
    printf("%d\n",ans);
    return 0;
}