[SHOI 2007] 善意的投票

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=1934

[算法]

        首先 ， 选择睡觉的人和不选择睡觉的人构成两个集合

        这启发我们用最小割解决该问题 :

        1. 将源点与每个睡觉的人连边 ， 将每个不睡觉的人与汇点连边 ， 割掉这样的一条边的含义是 ： 有一个人放弃了睡觉 / 不睡觉 ， 产生了1冲突

        2. 将朋友之间连边 ， 格调这样一条边的含义是 ： 这两个人产生了冲突

        求解这个图的最小割即可

         时间复杂度 : O(dinic(N , M))

[代码]

#include<bits/stdc++.h>
using namespace std;
#define N 310
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int inf = 1e9;

struct edge
{
    int to , w , nxt;
} e[N * N * 4];

int tot , n , m , S , T;
int head[N] , dep[N];

template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(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 s)
{
    queue< int > q;
    memset(dep , 255 , sizeof(dep));
    q.push(s);
    dep[s] = 1;
    while (!q.empty())
    {
        int cur = q.front();
        q.pop();
        for (int i = head[cur]; i; i = e[i].nxt)
        {
            int v = e[i].to , w = e[i].w;
            if (w > 0 && dep[v] == -1)
            {
                dep[v] = dep[cur] + 1;
                q.push(v);
                if (v == T) return true;
            }
        }
    }
    return false;
}
inline int dinic(int u , int flow)
{
    int rest = flow;
    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 (w > 0 && dep[v] == dep[u] + 1)
        {
            int k = dinic(v , min(rest , w));
            if (!k) dep[v] = 0;
            rest -= k;
            e[i].w -= k;
            e[i ^ 1].w += k;
        }
    }
    return flow - rest;
}
 
int main()
{
    
    read(n); read(m);
    tot = 1;
    S = n + 1 , T = S + 1;
    for (int i = 1; i <= n; i++)
    {
        int x;
        read(x);
        if (!x) addedge(S , i , 1);
        else addedge(i , T , 1);
    }
    for (int i = 1; i <= m; i++)
    {
        int x , y;
        read(x); read(y);
        addedge(x , y , 1);
        addedge(y , x , 1);
    }
    int ans = 0;
    while (bfs(S))
    {
        while (int flow = dinic(S , inf)) ans += flow;
    }
    printf("%d\n" , ans);
    
    return 0;
}

标签：善意,long,return,int,flow,rest,dep,2007,SHOI
来源： https://www.cnblogs.com/evenbao/p/10355663.html