[Usaco2009 MAR] Earthquake Damage 2

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=1585

[算法]

        一个最小割的经典模型 ， 详见代码

        时间复杂度 : O(dinic(2N , 2C))

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 8010
#define MAXC 50010
const int inf = 2e9;

struct edge
{
        int to , w , nxt;
} e[MAXC << 1];

int tot , n , c , p , S , T;
int a[MAXN] , depth[MAXN], head[MAXN];
 
template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u , int v , int w)
{
        ++tot;
        e[tot] = (edge){v , w , head[u]};
        head[u] = tot;
        ++tot;
        e[tot] = (edge){u , 0 , head[v]};
        head[v] = tot;
}
inline bool bfs()
{
        int l , r;
        static int q[MAXN];
        q[l = r = 1] = S;
        memset(depth , 0 ,sizeof(depth));
        depth[S] = 1;
        while (l <= r)
        {
                int cur = q[l++];
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w;
                        if (w > 0 && !depth[v])
                        {
                                depth[v] = depth[cur] + 1;
                                q[++r] = v;
                                if (v == T) return true;
                        }
                }
        }
        return false;
}
inline int dinic(int u , int flow)
{
        int rest = flow , ret = 0;
        if (u == T) return flow;
        for (int i = head[u]; i && rest; i = e[i].nxt)
        {
                int v = e[i].to , w = e[i].w;
                if (depth[v] == depth[u] + 1 && w)
                {
                        int k = dinic(v , min(rest , w));
                        if (!k) depth[v] = 0;    
                        e[i].w -= k;
                        e[i ^ 1].w += k;
                        rest -= k;
                }    
        }        
        return flow - rest;
}

int main()
{
        
        read(n); read(c); read(p);
        S = 2 * n + 1 , T = S + 1;
        tot = 1;
        addedge(S , 1 , inf);
        for (int i = 1; i <= c; i++)
        {
                int x , y;
                read(x); read(y);
                addedge(x + n , y , inf);
                addedge(y + n , x , inf);
        }
        for (int i = 1; i <= p; i++)
        {
                int x;
                read(x);
                a[x] = true;
        }
        addedge(1 , n + 1 , inf);
        for (int i = 2; i <= n; i++)
        {
                if (a[i]) 
                {
                        addedge(i , n + i , inf);
                        addedge(i , T , inf);
                } else addedge(i , n + i , 1);
        }
        int ans = 0;
        while (bfs())
        {
                while (int flow = dinic(S , inf)) ans += flow;        
        }
        printf("%d\n" , ans);
        return 0;
    
}