题解
直接二分然后建图跑网络流看看是否合法即可
就是源点向每个激光武器连一条二分到的时间×激光武器每秒攻击值的边
每个激光武器向能攻击的装甲连一条边
每个装甲向汇点连一条装甲值的边
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 30005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,M;
int A[55],B[55],g[55][55];
struct node {
int to,next;
db cap;
}E[1000005];
int sumE,head[205],cur[205],S,T,lev[205];
db all = 0;
void add(int u,int v,db c) {
E[++sumE].to = v;
E[sumE].cap = c;
E[sumE].next = head[u];
head[u] = sumE;
}
void addtwo(int u,int v,db c) {
add(u,v,c);add(v,u,0);
}
void Init() {
read(N);read(M);
for(int i = 1 ; i <= N ; ++i) {read(A[i]);all += A[i];}
for(int i = 1 ; i <= M ; ++i) read(B[i]);
for(int i = 1 ; i <= M ; ++i) {
for(int j = 1 ; j <= N ; ++j) {
read(g[i][j]);
}
}
}
bool BFS() {
static int que[205],ql,qr;
sumE = 0;
for(int i = S ; i <= T ; ++i) {cur[i] = head[i];lev[i] = -1;}
que[ql = qr = 1] = S;
lev[S] = 0;
while(ql <= qr) {
int u = que[ql++];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(E[i].cap > 0 && lev[v] == -1) {
lev[v] = lev[u] + 1;
que[++qr] = v;
if(v == T) return true;
}
}
}
return false;
}
db dfs(int u,db aug) {
if(u == T) return aug;
db flow = 0;
for(int &i = cur[u] ; i ; i = E[i].next) {
if(E[i].cap > 0) {
int e = i,v = E[i].to;
if(lev[v] > lev[u]) {
db t = dfs(v,min(E[e].cap,aug - flow));
flow += t;E[e].cap -= t;E[e ^ 1].cap += t;
if(flow == aug) break;
}
}
}
if(flow != aug) lev[u] = -1;
return flow;
}
bool check(db mid) {
sumE = 1;memset(head,0,sizeof(head));
S = 1,T = 1 + M + N + 1;
for(int i = 1 ; i <= M ; ++i) {
addtwo(S,S + i,mid * B[i]);
}
for(int i = 1 ; i <= N ; ++i) {
addtwo(1 + M + i,T,A[i]);
}
for(int i = 1 ; i <= M ; ++i) {
for(int j = 1 ; j <= N ; ++j) {
if(g[i][j]) addtwo(1 + i,1 + M + j,1000000000);
}
}
db res = 0;
while(BFS()) {
res += dfs(S,1000000000);
}
if(res < all) return false;
return true;
}
void Solve() {
db L = 0,R = 100000 * N;
int cnt = 60;
while(cnt--) {
db mid = (L + R) / 2;
if(check(mid)) R = mid;
else L = mid;
}
printf("%.3lf
",L);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}