bzoj 1070 [SCOI2007]修车 费用流
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题面
解法
感觉解法还是挺妙的
将每一个工人都拆成(n)个点,总共(n×m)个点
然后第(i)个工人的第(j)个点表示某辆车是在倒数第(j)个开始修的
然后就转化成修车的时间对总时间的贡献
连接(s)和(n)辆车,容量为1,费用为0,
连接(t)和最后的(n×m)个点,容量为1,费用为0
连接代表车的(n)个点和代表工人的(n×m)个点,容量为1,费用为(k×tim_{i,j})
然后跑一遍费用流即可
代码
#include <bits/stdc++.h>
#define N 1010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == ‘-‘) f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - ‘0‘, c = getchar(); x *= f;
}
struct Edge {
int next, num, c, w;
} e[N * N];
int s, t, cnt, dis[N], pre[N], las[N], used[N], tim[N][N];
void add(int x, int y, int c, int w) {
e[++cnt] = (Edge) {e[x].next, y, c, w};
e[x].next = cnt;
}
void Add(int x, int y, int c, int w) {
add(x, y, c, w), add(y, x, 0, -w);
}
bool spfa(int s, int t) {
queue <int> q; q.push(s);
for (int i = 0; i <= t; i++)
dis[i] = INT_MAX, used[i] = 0;
dis[s] = 0; used[s] = 1;
while (!q.empty()) {
int x = q.front(); used[x] = 0; q.pop();
for (int p = e[x].next; p; p = e[p].next) {
int k = e[p].num, c = e[p].c, w = e[p].w;
if (c && dis[k] > dis[x] + w) {
dis[k] = dis[x] + w;
pre[k] = x, las[k] = p;
if (!used[k]) q.push(k), used[k] = 1;
}
}
}
return dis[t] != INT_MAX;
}
int EK(int s, int t) {
int ret = 0;
while (spfa(s, t)) {
int fl = INT_MAX;
for (int x = t; x != s; x = pre[x])
chkmin(fl, e[las[x]].c);
ret += dis[t] * fl;
for (int x = t; x != s; x = pre[x])
e[las[x]].c -= fl, e[las[x] ^ 1].c += fl;
}
return ret;
}
int main() {
int n, m; read(n), read(m);
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
read(tim[i][j]);
s = 0, t = cnt = m + n * m + 1;
if (cnt % 2 == 0) cnt++;
for (int i = 1; i <= n * m; i++) Add(s, i, 1, 0);
for (int i = n * m + 1; i <= n * m + m; i++) Add(i, t, 1, 0);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int k = 1; k <= m; k++)
Add((i - 1) * m + j, n * m + k, 1, tim[k][i] * j);
cout << fixed << setprecision(2) << (double)EK(s, t) / m << "
";
return 0;
}
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bzoj 1070: [SCOI2007]修车 -- 费用流