[BZOJ 2127] Happiness

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=2127

[算法]

         首先默认每个人都选文科 

         那么 , "选"就是指选理科 , 而"不选"就是指选文科

         那么选所获得的收益就是(V理 - V文)

         而额外获得的收益可以看作是 : 若两个点同时选 , 可以获得一些收益和若两个点中有一个不选 , 则会失去收益

         解最大权闭合子图 , 即可

        时间复杂度 : O(dinic(N,M))

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define N 110
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int inf = 2e9;

struct edge
{
        int to , w , nxt;
} e[N * N * 50];

int total , tot , S , T , n , m;
int head[N * N * 5] , dep[N * N * 5] , a[N][N] , b[N][N] , point[N][N];

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()
{
        queue< int > q;
        for (int i = 1; i <= total; i++) dep[i] = -1;
        dep[T] = -1;
        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);
        int ans = 0;
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        read(a[i][j]);        
                }        
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        read(b[i][j]);
                        ans += b[i][j];
                }
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        point[i][j] = ++total;                
                }
        }
        tot = 1;
        S = 0 , T = 5 * N * N - 1;
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        if (a[i][j] >= b[i][j]) 
                        {
                                ans += a[i][j] - b[i][j];
                                addedge(S , point[i][j] , a[i][j] - b[i][j]);
                        } else addedge(point[i][j] , T  , b[i][j] - a[i][j]);        
                }        
        }
        // NM + 4NM = 5NM
        // 24NM + 2NM = 26NM
        for (int i = 1; i < n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        int x;
                        read(x);
                        ans += x;
                        ++total;
                        addedge(S , total , x);
                        addedge(total , point[i][j] , inf);
                        addedge(total , point[i + 1][j] , inf);
                }
        }
        for (int i = 1; i < n; i++)
        {
                for (int j = 1; j <= m; j++)
                {
                        int x;
                        read(x);
                        ans += x;
                        ++total;
                        addedge(total , T , x);
                        addedge(point[i][j] , total , inf);
                        addedge(point[i + 1][j] , total , inf);
                }
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j < m; j++)
                {
                        int x;
                        read(x);
                        ans += x;
                        ++total;
                        addedge(S , total , x);
                        addedge(total , point[i][j] , inf);
                        addedge(total , point[i][j + 1] , inf);
                }
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j < m; j++)
                {
                        int x;
                        read(x);
                        ans += x;
                        ++total;
                        addedge(total , T , x);
                        addedge(point[i][j] , total , inf);
                        addedge(point[i][j + 1] , total , inf);
                }
        }
        while (bfs())
        {
                while (int flow = dinic(S , inf)) ans -= flow;
        }
        printf("%d\n" , ans);
        
        return 0;
    
}

 

         

posted @ 2019-02-10 21:19  evenbao  阅读(179)  评论(0编辑  收藏  举报