Codechef 14.12 RIN 网络流+最小割
题目大意：
有$n$个课程，$m$个学期，每个课程都必须选择在某个学期学习，第$i$个课程在第$j$个学期学习将会得到$X_{i,j}$的价值。同时有$k$个限制条件，第$i$个限制条件给定$A_{i},B_{i}$，要求课程$A_{i}$所在的学期必须在$B_{i}$所在的学期之前。
求所有课程价值的最大平均值。
数据范围$n,m,k\leq{100}$，$0\leq{X_{i,j}}\leq{100}$，保证至少存在一种合法的解。
算法讨论：
将每个课程拆成一条长度为$m+1$的链，分别表示第$0-m$个学期，对于课程$i$，第$j-1(1\leq{j}\leq{m})$个点到第$j$个点连一条边，容量为$100-X_{i,j}$。
对于源点，连向所有链的第$0$个学期，容量为$\infty$。
所有链的第$m$个学期连向汇点，容量为$\infty$。
对于限制$A,B$，对于$1\leq{i}\leq{m}$，从$A$链的第$i-1$个学期连向$B$链的第$i$个学期，容量为$\infty$，这样就能满足题目中的限制了。
于是我们求最小割，再用$100\times{n}$减去就是最大的总和了。
再除以$n$就是最大平均值。
时空复杂度：
时间复杂度$O(MaxFlow(n\times{m},(n+k)\times{m}))$，空间复杂度$O((n+k)\times{m})$。
代码：
#include <cstdio>
#include <cstring>
#include <cctype>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f
struct Solver {
static const int V = 100010;
static const int E = 1000010;
int head[V], nxt[E], flow[E], to[E], ind, d[V];
void reset() {
ind = 0;
memset(head, -1, sizeof head);
}
void addedge(int a, int b, int c) {
int q = ind++;
to[q] = b;
nxt[q] = head[a];
head[a] = q;
flow[q] = c;
}
void make(int a, int b, int c) {
addedge(a, b, c);
addedge(b, a, 0);
}
bool bfs(int s, int t, int l, int r) {
static int q[V], fr, ta, i, j;
for (i = l; i <= r; ++i)
d[i] = -1;
fr = ta = 0;
d[q[ta++] = s] = 0;
while (fr != ta) {
i = q[fr++];
for (j = head[i]; j != -1; j = nxt[j])
if (flow[j] && d[to[j]] == -1)
d[q[ta++] = to[j]] = d[i] + 1;
}
return d[t] != -1;
}
int dinic(int p, int t, int Maxflow) {
if (p == t)
return Maxflow;
int last = Maxflow;
for (int j = head[p]; j != -1; j = nxt[j]) {
if (flow[j] && d[to[j]] == d[p] + 1) {
int toflow = dinic(to[j], t, last > flow[j] ? flow[j] : last);
if (toflow) {
flow[j] -= toflow;
flow[j ^ 1] += toflow;
if (!(last -= toflow))
return Maxflow;
}
}
}
d[p] = -1;
return Maxflow - last;
}
int Maxflow(int s, int t, int l, int r) {
int ans = 0;
while (bfs(s, t, l, r))
ans += dinic(s, t, inf);
return ans;
}
}g;
int node[110][110];
int main() {
int n, m, k, x, a, b;
cin >> n >> m >> k;
int i, j, id = 0;
for (i = 1; i <= n; ++i)
for (j = 0; j <= m; ++j)
node[i][j] = ++id;
g.reset();
for (i = 1; i <= n; ++i) {
for (j = 1; j <= m; ++j) {
cin >> x;
if (x == -1)
g.make(node[i][j - 1], node[i][j], inf);
else
g.make(node[i][j - 1], node[i][j], 100 - x);
}
}
int s = 0, t = ++id;
for (i = 1; i <= n; ++i) {
g.make(s, node[i][0], inf);
g.make(node[i][m], t, inf);
}
while (k--) {
cin >> a >> b;
for (i = 1; i <= m; ++i)
g.make(node[a][i - 1], node[b][i], inf);
}
printf("%.2lf\n", (n * 100 - g.Maxflow(s, t, 0, id)) / (double)n);
return 0;
}
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