[BZOJ2127]happiness
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;
}
