In order to learn ISAP have done this problem. Feel ISAP and no more than dinic difficult to hit how much ...
Second, the source point to the laser edge, the weight of the total damage, laser to the robot can reach infinity, the robot to the meeting point, the weight of the armor value. Then meeting Point's Bender satisfies the answer.
But the strange thing is to set the upper bound of the second point to 5e6, and change it to sum (all the armor values of the robots). Then add a sentence of if (l>sum) while (1); After that, it's over. The answer will not exceed sum ah ... I hope the old driver can take me with me.
The AC code is as follows:
#include <iostream> #include <cstdio> #include <cstring> #define N #define M 5505 #define INF 10000000
using namespace Std;
int n,m,tot=1,gol,fst[n],cur[n],pnt[m],nxt[m],d[n],num[n],h[n],fa[n];
Double sum,a[n],b[n],len[m]; void Add (int x,int y,double z) {pnt[++tot]=y; len[tot]=z; nxt[tot]=fst[x]; fst[x]=tot;} void BFs () {memset (num,0,sizeof
(num)); Memset (d,0x3f,sizeof (d));
d[gol]=0; int head=0,tail=1;
H[1]=gol;
while (head<tail) {int x=h[++head],p; num[d[x]]++;
for (p=fst[x]; p. p=nxt[p]) if (len[p^1]) {int y=pnt[p];
if (d[y]>inf) {d[y]=d[x]+1; h[++tail]=y;
Double Up () {int i; double tmp=inf;
for (I=gol i; i=pnt[fa[i]^1]) tmp=min (Tmp,len[fa[i]);
for (I=gol i; i=pnt[fa[i]^1]) {len[fa[i]]-=tmp; len[fa[i]^1]+=tmp;
return TMP;
bool OK (double t) {int p,i,x=0;
for (i=2; i<=tot; i+=2) {len[i]+=len[i^1]; len[i^1]=0;
for (p=fst[0); p. p=nxt[p]) len[p]=b[pnt[p]]*t;
memcpy (cur,fst,sizeof (FST)); Double flow=0; BFS ();
while (D[0]<=gol) {if (X==gol) {flow+=up (); x=0;}
BOOL Flag=1;
for (p=cur[x]; p. p=nxt[p]) if (len[p]) {int y=pnt[p];
if (d[x]==d[y]+1) {cur[x]=fa[y]=p; x=y; flag=0;
Break
} if (flag) {int mn=gol; cur[x]=fst[x];
for (p=fst[x]; p; p=nxt[p]) if (len[p)) Mn=min (Mn,d[pnt[p]); num[d[x]]--;
if (!num[d[x]]) break; d[x]=mn+1;
num[d[x]]++;
if (x) x=pnt[fa[x]^1];
} return flow>=sum;
int main () {scanf ("%d%d", &n,&m); gol=n+m+1; int i,j,x;
For (I=1 i<=n; i++) {scanf ("%lf", &a[i]); sum+=a[i]; Add (I+m,gol,a[i]);
Add (gol,i+m,0);
For (I=1 i<=m; i++) {scanf ("%lf", &b[i]), add (0,i,b[i)), add (i,0,0);
For (I=1 i<=m; i++) for (j=1; j<=n; J + +) {scanf ("%d", &x);
if (x) {Add (I,j+m,inf); add (j+m,i,0);}
Double L=0,r=5e6,mid;
while (L+1e-4<r) {mid= (l+r)/2; if (OK (mid)) R=mid;
else L=mid;
printf ("%.6f\n", L);
return 0;
}
by Lych
2016.4.9