[BZOJ1070]修车

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题目链接：BZOJ1070

题目大意
啊，好懒啊，中文题面以后就不说了吧。

分析
1. 感觉上这是一道网络流的题，却想不出来怎么建图；其实这道题的建图思路很巧妙。
2. 对于一个工作人员i，假设他修的倒数第k辆车是j，那么会导致总等待时间增加k×cost[i][j]，这样以来，其花费就之和k与cost[i][j]有关了。
3. 把每个工作人员拆成n个点X[i][k]，向所有车子Y[j]连边，容量为1，花费为k×cost[i][j]代表其在倒数第k辆的时候修第j辆车。
4. S向所有X连边，容量为1，费用为0；所有Y向T连边，容量为1，费用为0。
5. 跑一遍费用流，答案除以n就好了。

上代码

#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 10 + 5;const int M = 60 + 5;const int INF = 0x3f3f3f3f;int n, m;int S, T, cnt;int C[N][M], id1[N][M], id2[M];int head[N * M], len;struct nodeLib {    int to, nxt, val, flow;    inline void add(int a, int b, int c, int d) {        to = b, val = d, flow = c;        nxt = head[a], head[a] = len++;    }} lib[N * M * M << 1];inline int read() {    char ch;    int ans = 0, neg = 1;    while (ch = getchar(), ch < '0' || ch > '9')        if (ch == '-') neg = -1;    while (ch >= '0' && ch <= '9')        ans = ans * 10 + ch - '0', ch = getchar();    return ans * neg;}inline void makePath(int a, int b, int c, int d) {    lib[len].add(a, b, c, d);    lib[len].add(b, a, 0, -d);}void init() {    n = read(), m = read();    len = 2, S = ++cnt, T = ++cnt;    for (int j = 1; j <= m; j++)        for (int i = 1; i <= n; i++)            C[i][j] = read();    for (int i = 1; i <= n; i++)        for (int j = 1; j <= m; j++)            makePath(S, id1[i][j] = ++cnt, 1, 0);    for (int i = 1; i <= m; i++)        makePath(id2[i] = ++cnt, T, 1, 0);    for (int i = 1; i <= n; i++)        for (int j = 1; j <= m; j++)            for (int k = 1; k <= m; k++)                makePath(id1[i][j], id2[k], 1, j * C[i][k]);}queue <int> Q;bool inQ[N * M];int dist[N * M], preE[N * M], preV[N * M];bool SPFA() {    memset(dist, 0x3f, sizeof(dist));    dist[S] = 0, Q.push(S), inQ[S] = true;    while (!Q.empty()) {        int tmp = Q.front();        Q.pop(), inQ[tmp] = false;        for (int p = head[tmp]; p; p = lib[p].nxt) {            int now = lib[p].to, val = lib[p].val;            if (lib[p].flow && dist[now] > dist[tmp] + val) {                dist[now] = dist[tmp] + val;                preE[now] = p, preV[now] = tmp;                if (!inQ[now]) Q.push(now), inQ[now] = true;            }        }    }    return dist[T] != INF;}int MCMF() {    int ans = 0;    while (SPFA()) {        int maxf = INF;        for (int p = T; p != S; p = preV[p])            maxf = min(maxf, lib[preE[p]].flow);        for (int p = T; p != S; p = preV[p])            lib[preE[p]].flow -= maxf, lib[preE[p] ^ 1].flow += maxf;        ans += dist[T] * maxf;    }    return ans;}int main() {    init();    printf("%.2lf\n", (double)MCMF() / m);    return 0;}

我还是naive啊，没有想出来建图。
以上

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