这个题太恶心了。。。并不想继续做了。。。
本代码在bzoj上TLE!
大致说一下思路: 建立ST,首先由S连边(S,u,a)a代表学文的分数,连向T(u,T,b)b表示学理的分数,这样构造出了两个人独立的分数。
然后考虑联合分数,对于相邻的两个点xy,看下图(盗个图:
设xy都学文的分数为w1,都学理的分数为w2,则a=w1/2,b=w1/2,c=w2/2,d=w2/2,e=(w1+w2)/2,每一种割与其对应的亏损分数如下:
a+b -w1 都学理->w2
c+d -w2 都学文->w1
a+d+e -w1-w2 不同-> 0
c+d+e -w1-w2 ...
注意双向边e,我们是变成两条有向边加入网络,而又因为我们求最小割用的是最大流的算法,所以这条边可以看作是一条双向且权值为e的边。
然后把权值*2,解决精度问题。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 105; /////////////////////////////////////////////
const int maxv = maxn * maxn * 2;
const int inf = INT_MAX;
int n, m, s, t, v;
struct edge {
int from;
int to;
int cap;
};
vector<edge> edges;
vector<int> G[maxv];
int dist[maxv], iter[maxv];
int z[maxv][maxv];
bool zz[maxv][maxv];
inline int read() {
char c = getchar();
int f = 1, x = 0;
while (!isdigit(c)) {
if (c == '-')
f = -1;
c = getchar();
}
while (isdigit(c))
x = x * 10 + c - '0', c = getchar();
return x * f;
}
inline void add_edge(int from, int to, int cap) {
edges.push_back((edge){from, to, cap});
edges.push_back((edge){to, from, 0});
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
inline void bfs(int s) {
memset(dist, -1, sizeof(dist));
dist[s] = 0;
queue<int> q;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = 0; i < G[u].size(); i++) {
edge &e = edges[G[u][i]];
if (e.cap > 0 && dist[e.to] == -1) {
dist[e.to] = dist[u] + 1;
q.push(e.to);
}
}
}
}
inline int dfs(int s, int t, int flow) {
if (s == t)
return flow;
for (int &i = iter[s]; i < G[s].size(); i++) {
edge &e = edges[G[s][i]];
if (e.cap > 0 && dist[e.to] > dist[s]) {
int d = dfs(e.to, t, min(flow, e.cap));
if (d > 0) {
e.cap -= d;
edges[G[s][i] ^ 1].cap += d;
return d;
}
}
}
return 0;
}
inline int dinic(int s, int t) {
int flow = 0;
while (1) {
bfs(s);
if (dist[t] == -1)
return flow;
memset(iter, 0, sizeof(iter));
int F;
while (F = dfs(s, t, inf))
flow += F;
}
}
int main() {
// freopen("input", "r", stdin); //////////////////////////
memset(z, 0, sizeof(z));
memset(zz, 0, sizeof(zz));
scanf("%d %d", &n, &m);
s = 0; //文科
t = n * m + 1; //理科
v = t + 1;
int ans = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
int x;
x = read();
x <<= 1;
z[s][(i - 1) * m + j] += x;
zz[s][(i - 1) * m + j] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
int x;
x = read();
x <<= 1;
z[(i - 1) * m + j][t] += x;
zz[(i - 1) * m + j][t] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j <= m; j++) {
int x;
x = read();
z[(i - 1) * m + j][i * m + j] += x;
z[s][(i - 1) * m + j] += x;
z[s][i * m + j] += x;
zz[(i - 1) * m + j][i * m + j] = 1;
zz[s][(i - 1) * m + j] = 1;
zz[s][i * m + j] = 1;
add_edge((i - 1) * m + j, i * m + j, x);
add_edge(i * m + j, (i - 1) * m + j, x);
}
}
for (int i = 1; i < n; i++) {
for (int j = 1; j <= m; j++) {
int x;
x = read();
z[(i - 1) * m + j][i * m + j] += x;
z[(i - 1) * m + j][t] += x;
z[i * m + j][t] += x;
zz[(i - 1) * m + j][i * m + j] = 1;
zz[(i - 1) * m + j][t] = 1;
zz[i * m + j][t] = 1;
add_edge((i - 1) * m + j, i * m + j, x);
add_edge(i * m + j, (i - 1) * m + j, x);
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j < m; j++) {
int x;
x = read();
z[(i - 1) * m + j][(i - 1) * m + j + 1] += x;
z[s][(i - 1) * m + j] += x;
z[s][(i - 1) * m + j + 1] += x;
zz[(i - 1) * m + j][(i - 1) * m + j + 1] = 1;
zz[s][(i - 1) * m + j] = 1;
zz[s][(i - 1) * m + j + 1] = 1;
add_edge((i - 1) * m + j, (i - 1) * m + j + 1, x);
add_edge((i - 1) * m + j + 1, (i - 1) * m + j, x);
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j < m; j++) {
int x;
x = read();
z[(i - 1) * m + j][(i - 1) * m + j + 1] += x;
z[(i - 1) * m + j][t] += x;
z[(i - 1) * m + j + 1][t] += x;
zz[(i - 1) * m + j][(i - 1) * m + j + 1] = 1;
zz[(i - 1) * m + j][t] = 1;
zz[(i - 1) * m + j + 1][t] = 1;
add_edge((i - 1) * m + j, (i - 1) * m + j + 1, x);
add_edge((i - 1) * m + j + 1, (i - 1) * m + j, x);
}
}
for (int i = 1; i < t; i++) {
ans += z[s][i];
ans += z[i][t];
if (zz[s][i])
add_edge(s, i, z[s][i]);
if (zz[i][t])
add_edge(i, t, z[i][t]);
}
for (int i = 1; i < t; i++) {
for (int j = 1; j < t; j++) {
if (zz[i][j]) {
add_edge(i, j, z[i][j]);
add_edge(j, i, z[i][j]);
}
}
}
/* for (int i = 0; i < v; i++) {
cout << "For " << i << ':' << endl;
for (int j = 0; j < G[i].size(); j++) {
edge &e = edges[G[i][j]];
if (e.cap > 0)
cout << "to " << e.to << " cap " << e.cap << endl;
}
}*/
ans -= dinic(s, t);
printf("%d
", ans >> 1);
}