[BZOJ2127]happiness

题面戳我

Solution

一般这种题就转化成最小割做
把最大收益转化成最小损失,先把所有收益加入ans

考虑建图,设S集合为选文的,T为选理的
单个选的比较简单,就直接连就好了:
直接令容量(S,x)=选文科的收益,(x,T)=选理科的收益即可。
那么两个一起选的怎么连?
设两个人x,y,他们俩一起选文的收益是a,一起选理的收益是b。
四种情况

  • x,y都选文。此时被割掉的边是(x,T)和(y,T),总损失应为b
  • x,y都选理。此时被割掉的边是(S,x)和(S,y),总损失应为a
  • x选文y选理。此时被割掉的边是(x,T),(S,y)和(x,y),总损失应为a + b
  • x选理y选文。此时被割掉的边是(S,x),(y,T)和(y,x),总损失应为a + b

这样就能列出一个方程组:

\[\begin{cases} (x,T) + (y, T) = b\\ (S, x) + (S, y) = a\\ (x, T) + (S, y) + (x, y) = a + b\\ (S, x) + (y, T) + (y, x) = a + b\\ \end{cases} \]

显然无解,这是不定方程
考虑到不影响其它边,那么取任意一组可行解就好

\[(x, T) = \frac {b}{2}\\ (y, T) = \frac {b}{2}\\ (S, x) = \frac {a}{2}\\ (S, y) = \frac {a}{2}\\ (x, y) = \frac {a + b}{2}\\ (y, x) = \frac {a + b}{2}\\ \]

连完跑最大流即可

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(100010), __(1e6 + 10), INF(2147483647);

IL ll Read(){
    RG char c = getchar(); RG ll x = 0, z = 1;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, ans, id[110][110], num, val[6][110][110];
int w[__], fst[_], nxt[__], to[__], cnt, S, T, lev[_], cur[_];
queue <int> Q;

IL void Add(RG int u, RG int v, RG int f, RG int _f){
    w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
    w[cnt] = _f; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}

IL int Dfs(RG int u, RG int maxf){
    if(u == T) return maxf;
    RG int ret = 0;
    for(RG int &e = cur[u]; e != -1; e = nxt[e]){
        if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
        RG int f = Dfs(to[e], min(w[e], maxf - ret));
        ret += f; w[e ^ 1] += f; w[e] -= f;
        if(ret == maxf) break;
    }
    return ret;
}

IL bool Bfs(){
    Fill(lev, 0); lev[S] = 1; Q.push(S);
    while(!Q.empty()){
        RG int u = Q.front(); Q.pop();
        for(RG int e = fst[u]; e != -1; e = nxt[e]){
            if(lev[to[e]] || !w[e]) continue;
            lev[to[e]] = lev[u] + 1;
            Q.push(to[e]);
        }
    }
    return lev[T];
}

int main(RG int argc, RG char* argv[]){
	n = Read(); m = Read(); Fill(fst, -1); T = n * m + 1;
	for(RG int i = 1; i <= n; ++i)
		for(RG int j = 1; j <= m; ++j)
			id[i][j] = ++num;
	for(RG int p = 0; p < 6; ++p){
		RG int x = n - ((p == 2) | (p == 3)), y = m - ((p == 4) | (p == 5));
		for(RG int i = 1; i <= x; ++i)
			for(RG int j = 1; j <= y; ++j){
				val[p][i][j] = Read(), ans += val[p][i][j];
				if(p == 0) Add(S, id[i][j], val[p][i][j] << 1, 0);
				if(p == 1) Add(id[i][j], T, val[p][i][j] << 1, 0);
				if(p == 3){
					RG int xx = id[i][j], yy = id[i + 1][j], b = val[p][i][j], a = val[p - 1][i][j];
					Add(xx, T, b, 0); Add(yy, T, b, 0);
					Add(S, xx, a, 0); Add(S, yy, a, 0);
					Add(xx, yy, a + b, a + b);
				}
				if(p == 5){
					RG int xx = id[i][j], yy = id[i][j + 1], b = val[p][i][j], a = val[p - 1][i][j];
					Add(xx, T, b, 0); Add(yy, T, b, 0);
					Add(S, xx, a, 0); Add(S, yy, a, 0);
					Add(xx, yy, a + b, a + b);
				}
			}
	}
    for(ans <<= 1; Bfs(); ) Copy(cur, fst), ans -= Dfs(S, INF);
	printf("%d\n", ans >> 1);
    return 0;
}

特别鸣谢ZSY大佬的教导

posted @ 2018-01-08 15:21  Cyhlnj  阅读(102)  评论(0编辑  收藏  举报