bzoj3993 [SDOI2015] Star Wars binary + maximum flow

Description

The X Corps and the Y Corps are fighting fiercely in 3333 on a planet in the Milky Way. At some stage of the fight, the Y Corps dispatched n massive robots to attack the X Regiment's position, of which the armor value of the first giant robot was AI. When the armor of a giant robot is reduced to 0 or less, the giant robot is destroyed. The X Legion has m laser weapons, of which the first laser weapon can cut the armor value of a giant robot bi per second. The attack of laser weapons is continuous. This laser weapon is so strange that a laser weapon can only attack certain enemies. The Y Corps saw their giant robots wiped out by one of the X legions, and they desperately needed more orders. For this purpose, the Y Corps needs to know how long it will take at least for the X Legion to destroy all the giant robots of the Y Corps. But they don't count the problem, so they turn to you for help. Input

First row, two integers, N, M.

Second row, n integers, A1, A2 ... An.
Third line, M integer, B1, B2 ... Bm.
The next m rows, n integers per row, these integers are either 0 or 1. The number of J integers in line I of this section is 0, which means that the first laser weapon is not allowed to attack the J Giant Robot, and 1 indicates that the first laser weapon can attack the first J giant Robot. Output

One line, a real number, represents the minimum amount of time that the X Legion will need to destroy all the giant robots in the Y Corps. The output result and the absolute error of the standard answer are not more than 10-3, which is considered correct. Sample Input

2 2

3

4 6 0

1 1

1

Sample Output

1.300000

HINT

"Sample Illustration 1"

After the battle began in the first 0.5 seconds, laser weapons 1 attack 2nd giant robots, laser weapons 2 attack the giant robot 1th. The 1th giant robot was completely destroyed, and the 2nd mega-robot still had 8 armor remaining;

For the next 0.8 seconds, laser weapons 1 and 2 attack the giant robot 2nd at the same time. The 2nd giant robot was completely destroyed.

For all the data, 1<=n, m<=50,1<=ai<=105,1<=bi<=1000, enter data to ensure that the X Legion will destroy all the giant droid Source of the Y Corps.

Round 1 Thanks yts1999 upload

Brush the water question edify sentiment ...

Note the upper bound of the two-point ... 1 billion will t to die ...

Deep don't turn into a bool again ...

#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <

Cmath> #include <queue> using namespace std;
typedef long Long LL;
const int SZ = 1010;
Const double INF = 5e6;

Const double EPS = 1e-4;

int head[sz],nxt[10000],tot = 1,n,m;
    struct edge{int t;
Double D;

}L[10000];
    void build (int f,int t,double d) {l[++ tot].t = t;
    L[TOT].D = D;
    Nxt[tot] = head[f];
HEAD[F] = tot;

} void Insert (int f,int t,double D) {build (F,t,d), build (t,f,0.0);
int Deep[sz];

Queue<int> Q;
    BOOL BFS (int s,int e) {memset (deep,0,sizeof (deep));
    while (Q.size ()) Q.pop ();
    Deep[s] = 1;
    Q.push (s);
        while (Q.size ()) {int u = q.front (); Q.pop ();
            for (int i = Head[u];i;i = Nxt[i]) {int v = l[i].t;
                if (l[i].d &&!deep[v]) {Deep[v] = Deep[u] + 1;
                if (v = = e) return true;
    Q.push (v);        }} return false;
    Double dfs (int u,double flow,int e) {if (Flow < EPS | | u = e) return flow;
    Double rest = flow;
        for (int i = Head[u];i;i = Nxt[i]) {int v = l[i].t;
            if (l[i].d && deep[v] = = Deep[u] + 1) {Double F = DFS (V,min (REST,L[I].D), E);
                if (F > EPs) {l[i].d = f;
                l[i ^ 1].d = f;
                rest = f;
            if (rest < EPS) break;
        else deep[v] = 0;
} return flow-rest;
    Double dinic (int s,int e) {double ans = 0;
    while (BFS (s,e)) ans + = DFS (s,inf,e);
return ans;
    } void Init () {memset head,0,sizeof (head);
tot = 1;

int A[sz],b[sz];

BOOL maps[233][233];
    BOOL Check (double mid) {init ();
    int s = n + M + 1;
    int e = n + M + 2;
    for (int i = 1;i <= m;i + +) Insert (s,i + n,mid * b[i));
    Double sum = 0; for (int i= 1;i <= n;i + +) Insert (I,e,a[i]), sum + = a[i]; for (int i = 1;i <= m;i + +) for (int j = 1;j <= N;j + +) if (maps[i][j)) Insert (i +
    N,j,mid * B[i]);
    Double ans = dinic (s,e);
return Fabs (Ans-sum) < EPS;
    Double Div () {double L = 0,r = INF;
        while (Fabs (r-l) > EPS) {double mid = (L + r)/2;
        if (check (mid)) R = Mid;
    else L = mid;
} return R;
    int main () {scanf ("%d%d", &n,&m);
    for (int i = 1;i <= n;i + +) scanf ("%d", &a[i]);
    for (int i = 1;i <= m;i + +) scanf ("%d", &b[i]);
    for (int i = 1;i <= m;i + +) for (int j = 1;j <= n;j + +) scanf ("%d", &maps[i][j]);
    printf ("%.6lf\n", div ());   
return 0;

 }