[Bzoj] 1066: [scoi2007] lizard (maximum Stream)

 

I should have understood this question.

We thought, each point has a limit, and each point can jump to another restricted point. Each point with a lizard can jump to a restricted point around, haha, naturally, you will think of network streams.

It is naturally possible to indicate the capacity limit of each point, so split the point, one up, one down, the capacity is the weight.

Connect a point in the distance to a point in the four-week connection. The capacity is oo.

Connects the source to the lizard with a capacity of 1

It can be connected to a downstream sink at a point outside the boundary, with a capacity of OO

When you run the largest stream at a time, the answer is the total number of bots minus the maximum flow.

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=1000, M=N*5000, oo=~0u>>1;int ihead[N], cnt=1, cur[N], gap[N], d[N], p[N], n, m, g, a[22][22];struct ED { int from, to, cap, next; } e[M];inline const int id(const int &x, const int &y) { return (x-1)*m+y; }inline const bool check(const int &i, const int &j, const int &x, const int &y) { return ((i-x)*(i-x)+(j-y)*(j-y))<=g*g; }inline void add(const int &u, const int &v, const int &w) {e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u; e[cnt].cap=w;e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v; e[cnt].cap=0;}int isap(const int &s, const int &t, const int &n) {for1(i, 0, n) cur[i]=ihead[i];int ret=0, i, f, u=s;gap[0]=n;while(d[s]<n) {for(i=cur[u]; i; i=e[i].next) if(e[i].cap && d[u]==d[e[i].to]+1) break;if(i) {p[e[i].to]=cur[u]=i; u=e[i].to;if(u==t) {for(f=oo; u!=s; u=e[p[u]].from) f=min(f, e[p[u]].cap);for(u=t; u!=s; u=e[p[u]].from) e[p[u]].cap-=f, e[p[u]^1].cap+=f;ret+=f;}}else {if(! (--gap[d[u]]) ) break;d[u]=n; cur[u]=ihead[u];for(i=ihead[u]; i; i=e[i].next) if(e[i].cap && d[u]>d[e[i].to]+1) d[u]=d[e[i].to]+1;++gap[d[u]];if(u!=s) u=e[p[u]].from;}}return ret;}int main() {read(n); read(m); read(g);int s=n*m*2+10, t=s+1, now, tp, ans=0; char c;for1(i, 1, n) for1(j, 1, m) {for(c=getchar(); c<‘0‘||c>‘9‘; c=getchar());tp=c-‘0‘;if(tp) {now=id(i, j); a[i][j]=tp;add(now, now+410, tp);}}for1(i, 1, n) for1(j, 1, m) {for(c=getchar(); c!=‘L‘&&c!=‘.‘; c=getchar());if(c==‘L‘) add(s, id(i, j), 1), ++ans;}for1(i, 1, n) for1(j, 1, m) if(a[i][j])for(int x=i-g; x<=i+g; ++x) for(int y=j-g; y<=j+g; ++y)if(a[x][y] && !(i==x && j==y) && check(i, j, x, y)) add(id(i, j)+410, id(x, y), oo);for1(i, 1, g) for1(j, 1, m) add(id(i, j)+410, t, oo), add(id(n-i+1, j)+410, t, oo);for1(i, 1, g) for1(j, 1, n) add(id(j, i)+410, t, oo), add(id(j, m-i+1)+410, t, oo);print(ans-isap(s, t, t+1));return 0;}

 

 

 

 

 

Description there are some different stone pillars in the grid map of column C in the r row, some of which are standing on some lizard. Your task is to escape as many lizard as possible out of the boundary. The distance between adjacent stone columns in each row is 1, and the LEAP distance of the lizard is D. That is, the lizard can jump to any stone column with a plane distance not greater than D. The stone column is unstable. When the lizard jumps every time, the height of the removed stone column is reduced by 1 (if it is still inside the map, the height of the stone column to be reached remains unchanged ), if the original height of the stone column is 1, the lizard disappears after it leaves. In the future, other lizard cannot settle. At any time, there cannot be two lizard on the same stone column. Input the first row of the input three integers: R, C, and D, that is, the distance between the map scale and the maximum hop. The initial status of the following R behavior bamboo, 0 indicates there is no stone column, 1 ~ 3 indicates the initial height of the stone column. The following R acts the position of the lizard. "L" indicates the lizard, and "." indicates that there is no lizard. Only one line of output contains an integer, that is, the minimum value of the total number of lizard that cannot be escaped. Sample input5 8 2
00000000
02000000
00321100
02000000
00000000
........
........
... Llll ..
........
........

Sample output1

Hint

100% of data meets the requirements: 1 <= r, C <= 20, 1 <= d <= 3

[Bzoj] 1066: [scoi2007] lizard (maximum Stream)