【LOJ】#2184. 「SDOI2015」星际战争

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('\n')
#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\n",L);
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}