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\n", ans);
    return 0;
}
View Code

  

posted @ 2019-07-27 10:14  Schenker  阅读(127)  评论(0编辑  收藏  举报