BZOJ1070 [SCOI2007]修车

蒟蒻表示不太会。。。

后来才想明白怎么做：一个拆了很多很多点的网络流模型。（想法其实是先考虑一个修理师的情况，再考虑多个的情况）

把每个修理师拆成n个点，一辆车分别想着n个点连边，容量为1，费用为t * x  其中x = 1, 2, 3 ... n，表示在这之后修理车的人要总共等待t, t * 2, t * 3, t * 4...时间

然后跑一边最小费用最大流就可以了。

看上去非常对，就是这么做。。。搞定！

（p.s. 作死小剧场：我建图的时候建错了，然后查spfa死活找不出错来>.<）

 

/**************************************************************
    Problem: 1070
    User: rausen
    Language: C++
    Result: Accepted
    Time:608 ms
    Memory:8744 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
const int inf = (int) 1e9;
struct edges{
    int next, to, cost, f;
} e[200000];
 
int m, n, tot = 1, g[10000], first[10000], d[10000], q[200000];
int a[1000][1000], s, t, ans;
bool v[10000];
 
inline void add_edge(int x, int y, int d, int c){
    e[++tot].next = first[x];
    first[x] = tot;
    e[tot].to = y;
    e[tot].f = d;
    e[tot].cost = c;
}
 
void add_Edges(int x, int y,int d, int c){
    add_edge(x, y, d, c);
    add_edge(y, x, 0, -c);
}
 
void build_graph(){
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            for(int k = 1; k <= n; ++k)
                add_Edges(i, j * n + k, 1, a[i][j] * k);
    s = m * n + n + 1,  t = m * n + n + 2;
    for (int i = 1; i <= n; ++i)
        add_Edges(s, i, 1, 0);
    for (int i = n + 1; i <= m * n + n; ++i)
        add_Edges(i, t, 1, 0);
}
 
inline int calc(){
    int flow = inf;
    for (int x = g[t]; x; x = g[e[x ^ 1].to])
        flow = min(flow, e[x].f);
    for (int x = g[t]; x; x = g[e[x ^ 1].to])
        e[x].f -= flow, e[x ^ 1].f += flow;
    return flow;
}
 
bool spfa(){
    for (int i = 1; i <= t; ++i)
        d[i] = inf;
    d[s] = 0, v[s] = 1;
    q[1] = s;
    int x, y;
    for(int l = 1, r = 1; l <= r; ++l){
        for (x = first[q[l]]; x; x = e[x].next){
            y = e[x].to;
            if (d[q[l]] + e[x].cost < d[y] && e[x].f){
                d[y] = d[q[l]] + e[x].cost;
                g[y] = x;
                if (!v[y])
                    q[++r] = y, v[y] = 1;
            }
        }
        v[q[l]] = 0;
    }
    return d[t] != inf;
}
 
int main(){
    scanf("%d%d", &m, &n);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            scanf("%d", a[i] + j);
     
    build_graph();
    while (spfa())
        ans += calc() * d[t];
    printf("%.2lf\n", (double) ans / n);
    return 0;
}