bzoj 1070 [SCOI2007]修车 费用流

题面

题目传送门

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 << "\n";
    return 0;
}