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  • Bzoj 2127 happiness 最小割

    happiness

    题解:

    将图转换成最小割.

    将割完的图中与S相连的点看做选文科, 与T相连的点看做选理科.

    flow(s, u) = 文科值

    flow(u,t) = 理科值

    假设u 和 v 一起选文科有奖励值z,  flow(s,u) = z/2  flow(s,v) = z/2, flow(u,v) = z/2

    假设u 和 v 一起选理科有奖励值z,  flow(u,t) = z/2  flow(v,t) = z/2, flow(u,v) = z/2

    然后合并边.

    具体理解可以画图看最小割之后的模型.

    剩下的图才是价值.

    #include<bits/stdc++.h>
    using namespace std;
    #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
    #define LL long long
    #define ULL unsigned LL
    #define fi first
    #define se second
    #define pb push_back
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define lch(x) tr[x].son[0]
    #define rch(x) tr[x].son[1]
    #define max3(a,b,c) max(a,max(b,c))
    #define min3(a,b,c) min(a,min(b,c))
    typedef pair<int,int> pll;
    const int inf = 0x3f3f3f3f;
    const int _inf = 0xc0c0c0c0;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const LL _INF = 0xc0c0c0c0c0c0c0c0;
    const LL mod =  (int)1e9+7;
    int a[6][105][105];
    int n, m;
    struct FFFlow{
        const static int N = 105*105;
        static const int M = N << 3;
        int head[N], deep[N], cur[N];
        int w[M], to[M], nx[M];
        int tot;
        void add(int u, int v, int val){
            w[tot]  = val; to[tot] = v;
            nx[tot] = head[u]; head[u] = tot++;
    
    //        w[tot] = 0; to[tot] = u;
    //        nx[tot] = head[v]; head[v] = tot++;
        }
        void Eadd(int u, int v, int val, int p){
            add(u, v, val);
            add(v, u, val * p);
        }
        int bfs(int s, int t){
            queue<int> q;
            memset(deep, 0, sizeof(deep));
            q.push(s);
            deep[s] = 1;
            while(!q.empty()){
                int u = q.front();
                q.pop();
                for(int i = head[u]; ~i; i = nx[i]){
                    if(w[i] > 0 && deep[to[i]] == 0){
                        deep[to[i]] = deep[u] + 1;
                        q.push(to[i]);
                    }
                }
            }
            return deep[t] > 0;
        }
        int Dfs(int u, int t, int flow){
            if(u == t) return flow;
            for(int &i = cur[u]; ~i; i = nx[i]){
                if(deep[u]+1 == deep[to[i]] && w[i] > 0){
                    int di = Dfs(to[i], t, min(w[i], flow));
                    if(di > 0){
                        w[i] -= di, w[i^1] += di;
                        return di;
                    }
                }
            }
            return 0;
        }
    
        int Dinic(int s, int t){
            int ans = 0, tmp;
            while(bfs(s, t)){
                for(int i = 0; i <= t; i++) cur[i] = head[i];
                while(tmp = Dfs(s, t, inf)) ans += tmp;
            }
            return ans;
        }
        void init(){
            memset(head, -1, sizeof(head));
            tot = 0;
        }
    
    }Flow;
    int main(){
        Flow.init();
        scanf("%d%d", &n, &m);
        int ans = 0;
        for(int k = 0; k < 6; ++k){
            int x = n, y = m;
            if(k == 2 || k == 3) --x;
            if(k == 5 || k == 4) --y;
            for(int i = 1; i <= x; ++i)
                for(int j = 1; j <= y; ++j){
                    scanf("%d", &a[k][i][j]);
                    ans += a[k][i][j];
                }
        }
        int s = 0, t = n * m + 1;
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                Flow.Eadd(s, (i-1)*m + j, a[0][i][j] * 2 + a[2][i-1][j] + a[2][i][j] + a[4][i][j-1] + a[4][i][j], 0);
                Flow.Eadd((i-1)*m + j, t, a[1][i][j] * 2 + a[3][i-1][j] + a[3][i][j] + a[5][i][j-1] + a[5][i][j], 0);
                if(i > 1) Flow.Eadd((i-1)*m+j, (i-2)*m+j, a[2][i-1][j] + a[3][i-1][j], 1);
                if(j > 1) Flow.Eadd((i-1)*m+j, (i-1)*m+j-1, a[4][i][j-1] + a[5][i][j-1], 1);
            }
        }
        ans -= Flow.Dinic(s, t) / 2;
        printf("%d
    ", ans);
        return 0;
    }
    View Code

      

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  • 原文地址:https://www.cnblogs.com/MingSD/p/11254013.html
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