BZOJ 3993: [SDOI2015]星际战争 [二分答案 二分图]

3993: [SDOI2015]星际战争

题意:略


R1D2T1考了裸二分答案+二分图最大匹配...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=105, M=1e5+5, INF=1e9+5;
const double eps=1e-12;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
} int n, m, s, t, a[N], b[N], g[55][55], sum;
struct edge{int v, ne; double c, f;}e[M];
int cnt=1, h[N];
inline void ins(int u, int v, double c) { //printf("ins %d %d %lf\n",u,v,c);
e[++cnt]=(edge){v, h[u], c, 0}; h[u]=cnt;
e[++cnt]=(edge){u, h[v], 0, 0}; h[v]=cnt;
}
int q[N], head, tail, d[N], vis[N];
bool bfs() { //puts("bfs");
memset(vis, 0, sizeof(vis));
head=tail=1;
q[tail++]=s; d[s]=0; vis[s]=1;
while(head != tail) {
int u=q[head++]; //printf("u %d\n",u);
for(int i=h[u];i;i=e[i].ne)
if(!vis[e[i].v] && e[i].c>e[i].f) {
vis[e[i].v]=1; d[e[i].v]=d[u]+1;
q[tail++]=e[i].v;
if(e[i].v == t) return true;
}
}
return false;
}
int cur[N];
double dfs(int u, double a) {
if(u==t || a==0) return a;
double flow=0, f;
for(int &i=cur[u];i;i=e[i].ne)
if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c-e[i].f)))>0 ) {
flow += f;
e[i].f += f;
e[i^1].f -= f;
a -= f;
if(a==0) break;
}
if(a) d[u]=-1;
return flow;
}
double dinic() {
double flow=0;
while(bfs()) {
for(int i=s; i<=t; i++) cur[i]=h[i];
flow += dfs(s, INF); //printf("flow %lf\n",flow);
}
//printf("flow %lf\n", flow);
return flow;
} bool check(double mid) {
cnt=1; memset(h, 0, sizeof(h));
for(int i=1; i<=m; i++) ins(s, i, b[i]*mid);
for(int i=1; i<=n; i++) ins(m+i, t, a[i]);
for(int i=1; i<=m; i++)
for(int j=1; j<=n; j++) if(g[i][j]) ins(i, m+j, INF); double flow=dinic();
return abs(flow-sum) < eps ? 1 : 0;
}
double l, r;
void solve() {
while(r-l > 1e-4) {
double mid=(l+r)/2.0;
if(check(mid)) r=mid;
else l=mid;
}
printf("%lf", l);
}
int main() {
freopen("in","r",stdin);
n=read(); m=read(); s=0; t=n+m+1;
for(int i=1; i<=n; i++) a[i]=read(), r+=a[i], sum+=a[i];
for(int i=1; i<=m; i++) b[i]=read();
for(int i=1; i<=m; i++) for(int j=1; j<=n; j++) g[i][j]=read();
solve();
}
上一篇:Loadrnner 参数化策略


下一篇:document事件及例子