【洛谷 P3705】 [SDOI2017]新生舞会（费用流，01分数规划）

题目链接
看到这题我想到了以前做过的一题，名字记不清了，反正里面有“矩阵”二字，然后是道二分图匹配的题。
经典的行列连边网络流。
第(i)行和第(j)列连边，费用为(b[i][j]-a[i][j] imes mid)，源点连行，列连汇点，跑最小费用最大流得到的最小费用取负，这个值就是最大的(sum a[i][j]-b[i][j] imes mid)，于是愉快的二分。
其实费用取反跑最小费用最大流再取反就是最大费用最大流。为什么不直接写最大费用最大流？我写WA了？反正就是错了。血的教训

#include <cstdio>
#include <queue>
#include <cstring>
#define INF 2147483647
using namespace std;
const int MAXN = 500;
const int MAXM = 200010;
const double eps = 1e-8;
queue <int> q; 
int s, t, now, n;
struct Edge{
    int from, next, to, rest;
    double cost;
}e[MAXM];
int head[MAXN], num = 1, vis[MAXN], Flow[MAXN], pre[MAXN], a[110][110], b[110][110], o;
double l, r, mid, dis[MAXN];
inline void Add(int from, int to, int flow, double cost){
    e[++num] = (Edge){ from, head[from], to, flow, cost }; head[from] = num;
    e[++num] = (Edge){ to, head[to], from, 0, -cost }; head[to] = num;
}
int RoadsExist(){
    q.push(s);
    for(int i = 1; i <= o; ++i)
       dis[i] = 1e10, pre[i] = 0;
    dis[s] = 0; Flow[s] = INF;
    while(!q.empty()){
      now = q.front(); q.pop(); vis[now] = 0;
      for(int i = head[now]; i; i = e[i].next)
         if(e[i].rest && dis[e[i].to] > dis[now] + e[i].cost){
           dis[e[i].to] = dis[now] + e[i].cost;
           pre[e[i].to] = i;
           Flow[e[i].to] = min(Flow[now], e[i].rest);
           if(!vis[e[i].to]){
             vis[e[i].to] = 1;
             q.push(e[i].to);
           }
         }
    }
    return pre[t];
}
double ek(){
	double maxcost = 0;
	while(RoadsExist()){
      maxcost += (double)Flow[t] * dis[t];
      for(int i = t; i != s; i = e[pre[i]].from){
         e[pre[i]].rest -= Flow[t];
         e[pre[i] ^ 1].rest += Flow[t];
      }
    }
    return -maxcost;
}
int main(){
    scanf("%d", &n); o = (n << 1) + 2; s = o - 1; t = o;
    for(int i = 1; i <= n; ++i)
       for(int j = 1; j <= n; ++j){
          scanf("%d", &a[i][j]);
          r += a[i][j];
       }
    for(int i = 1; i <= n; ++i)
       for(int j = 1; j <= n; ++j)
          scanf("%d", &b[i][j]);
    while(r - l > eps){
    	num = 1; mid = (l + r) / 2.0;
    	for(int i = 1; i <= o; ++i) head[i] = 0;
    	for(int i = 1; i <= n; ++i){
    	   Add(s, i, 1, 0), Add(i + n, t, 1, 0);
    	   for(int j = 1; j <= n; ++j)
    	      Add(i, j + n, 1, -(double)a[i][j] + (double)b[i][j] * mid);
    	}
    	if(ek() <= 0) r = mid;
    	else l = mid;
    }
    printf("%.6lf
", l);
    return 0;
}