BZOJ 3993 [SDOI2015]星际战争 | 网络流 二分答案

链接

BZOJ 3993

题解

这道题挺棵的……

二分答案t，然后源点向武器连t * b[i], 武器向能攻击的敌人连1, 敌人向汇点连a[i]，如果最大流等于所有敌人的a[i]之和则可行。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define enter putchar('\n')
#define space putchar(' ')
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c > '9' || c < '0')
	if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
	x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 233, M = 1000005, INF = 0x3f3f3f3f;
const double eps = 0.00001;
int n, m, src, des, sum, a[N], b[N];
bool conn[N][N];
int ecnt = 1, adj[N], cur[N], dis[N], go[M], nxt[M];
double cap[M];

void ADD(int u, int v, double _cap){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
    cap[ecnt] = _cap;
}
void add(int u, int v, double _cap){
    ADD(u, v, _cap);
    ADD(v, u, 0);
}
bool bfs(){
    static int que[N], qr;
    for(int i = 1; i <= des; i++)
	cur[i] = adj[i], dis[i] = -1;
    que[qr = 1] = src, dis[src] = 0;
    for(int ql = 1; ql <= qr; ql++){
	int u = que[ql];
	for(int e = adj[u], v; e; e = nxt[e])
	    if(cap[e] > 0 && dis[v = go[e]] == -1){
		dis[v] = dis[u] + 1, que[++qr] = v;
		if(v == des) return 1;
	    }
    }
    return 0;
}
double dfs(int u, double flow){
    if(u == des) return flow;
    double ret = 0, delta;
    for(int &e = cur[u], v; e; e = nxt[e])
	if(cap[e] > 0 && dis[v = go[e]] == dis[u] + 1){
	    delta = dfs(v, min(cap[e], flow - ret));
	    if(delta > 0){
		cap[e] -= delta;
		cap[e ^ 1] += delta;
		ret += delta;
		if(abs(flow - ret) < eps) return ret;
	    }
	}
    dis[u] = -1;
    return ret;
}
double maxflow(){
    double ret = 0;
    while(bfs()) ret += dfs(src, INF);
    return ret;
}
bool check(double mid){
    ecnt = 1;
    for(int i = 1; i <= des; i++)
	adj[i] = 0;
    for(int i = 1; i <= m; i++)
	add(src, i, mid * b[i]);
    for(int i = 1; i <= m; i++)
	for(int j = 1; j <= n; j++)
	    if(conn[i][j])
		add(i, j + m, INF);
    for(int i = 1; i <= n; i++)
	add(i + m, des, a[i]);
    return abs(maxflow() - sum) < eps;
}
int main(){

    read(n), read(m), src = n + m + 1, des = src + 1;
    for(int i = 1; i <= n; i++) read(a[i]), sum += 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(conn[i][j]);
    double l = 0, r = 100000, mid;
    while(r - l > 0.0001){
	mid = (l + r) / 2;
	if(check(mid)) r = mid;
	else l = mid;
    }
    printf("%lf\n", l);

    return 0;
}