"bzoj1066" [SCOI2007] Lizard network Max stream

"bzoj1066" [SCOI2007] Lizard description

In the grid map of a row C column, there are some highly different pillars, some of which stand on a few lizards, and your task is to get as many lizards as possible to escape the border. The distance of each column in each row is 1, and the lizard's jumping distance is D, that is, the lizard can jump to any one of the columns with a plane distance not exceeding d. The columns are unstable, and each time the lizard jumps, the height of the stone column is reduced by 1 (if it still falls inside the map, the height of the stone column is not changed), and if the column has a height of 1, the lizard disappears after it leaves. Other lizards cannot settle in the future. No two lizards can be found on the same stone column at any one time.

Input

Enter the first behavior three integer r,c,d, which is the size of the map and the maximum jump distance. The following r behavior is the initial state of the carnation, 0 means that there are no columns, and the initial height of the column. The following r behavior is the lizard position, "L" denotes the lizard, "." means no lizards.

Output

The output is only one row, containing an integer, the minimum number of lizards that cannot escape.

Sample INPUT5 8 2
00000000
02000000
00321100
02000000
00000000
........
........
.. Llll.
........
........ Sample Output1hint

100% data satisfies: 1<=r, c<=20, 1<=d<=3

Code

For each pillar, take the idea of dividing a point into two points (which can be abstracted from the bottom to the top), and its connection capacity is limited to the height of the stone column.

The super source is connected to all lizard-like points with a capacity of 1.

The super sink is connected to the bottom of all the points in the map that can jump out, and the capacity is INF.

For any of the two columns in the map, if the spacing is less than D, the bottom of one of the columns is connected to the top of the other pillar, and its connection capacity is INF.

The composition is complete, the rest is running the maximum flow, and then the number of lizards minus the maximum flow is the final result.

#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<queue>#include<typeinfo>#include<map>#include<stack>typedefLong Longll;using namespacestd;inline ll Read () {ll x=0, f=1; CharCh=GetChar ();  while(ch<'0'|| Ch>'9')    {        if(ch=='-') f=-1; CH=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; CH=GetChar (); }    returnx*F;}//***************************namespacenetflow{Const intmaxn=100000, maxm=500000, inf=1e9;structedge{intV,c,f,nx; Edge () {} Edge (intVintCintFintNX): V (v), C (c), F (f), NX (NX) {}} E[MAXM];intG[maxn],cur[maxn],pre[maxn],dis[maxn],gap[maxn],n,sz;voidInitint_n) {N=_n,sz=0; memset (G,-1,sizeof(g[0])*N);}voidLinkintUintVintc) {E[sz]=edge (V,c,0, G[u]); G[u]=sz++; E[SZ]=edge (U,0,0, G[v]); G[V]=sz++;}intISAP (intSintT) {    //T-S    intmaxflow=0, aug=inf,flag=false, U,v;  for(intI=0; i<n; ++i) cur[i]=g[i],gap[i]=dis[i]=0;  for(Gap[s]=n,u=pre[s]=s; dis[s]<n; flag=false)    {         for(int&it=cur[u]; ~it; it=e[it].nx) {            if(e[it].c>e[it].f&&dis[u]==dis[v=e[it].v]+1)            {                if(AUG&GT;E[IT].C-E[IT].F) aug=e[it].c-e[it].f; PRE[V]=u,u=v; Flag=true; if(u==T) { for(Maxflow+=aug; u!=S;) {E[cur[u=pre[u]]].f+=; E[cur[u]^1].f-=; } the=inf; }                 Break; }        }        if(flag)Continue; intmx=N;  for(intIt=g[u]; ~it; it=e[it].nx) {            if(e[it].c>e[it].f&&dis[e[it].v]<mx) {mx=DIS[E[IT].V]; Cur[u]=it; }        }        if((--gap[dis[u]) = =0) Break; ++gap[dis[u]=mx+1]; U=Pre[u]; }    returnMaxflow;}BOOLBFsintSintT) {    Static intQ[MAXN]; memset (DIS,-1,sizeof(dis[0])*N); Dis[s]=0; q[0]=S;  for(intH=0, t=1, U,v,it; h<t; ++h) { for(U=q[h],it=g[u]; ~it; it=e[it].nx) {            if(dis[v=e[it].v]==-1&&E[it].c>e[it].f) {Dis[v]=dis[u]+1; Q[t++]=v; }        }    }    returndis[t]!=-1;}intDfsintUintTintLow ) {    if(u==t)returnLow ; intret=0, Tmp,v;  for(int&it=cur[u]; ~it&&ret<low; it=e[it].nx) {        if(dis[v=e[it].v]==dis[u]+1&&E[it].c>e[it].f) {            if(Tmp=dfs (V,t,min (low-ret,e[it].c-E[IT].F)))  {ret+=tmp; E[IT].F+=tmp; E[it^1].f-=tmp; }        }    }    if(!ret) dis[u]=-1; returnret;}intDinic (intSintT) {    intmaxflow=0, TMP;  while(BFS (s,t)) {memcpy (cur,g,sizeof(g[0])*N);  while(Tmp=dfs (S,t,inf)) maxflow+=tmp; }    returnMaxflow;}}using namespaceNetFlow;DoubleDistintAintBintXinty) {    returnsqrt ((b-y) * (b-y) + (a-x) * (A-x));}intMain () {intn,m,d; Charmp[ -][ -],mp2[ -][ -]; CIN>>n>>m>>D; Init ( the);  for(intI=1; i<=n; i++) {scanf ("%s", mp[i]+1); }     for(intI=1; i<=n; i++) {scanf ("%s", mp2[i]+1); }     for(intI=1; i<=n; i++)    {         for(intj=1; j<=m; J + +)        {            if((mp[i][j]-'0') >0)            {                 for(inth=i-d; h<=i+d; h++)                {                     for(intk=j-d; k<=j+d; k++)                    {                        if(h==i&&j==k)Continue; if(h<0|| k<0|| h>n+1|| k>m+1)                            Continue; Doubledd=d*1.0; if(Dist (i,j,h,k) >dd)Continue; if(h==0|| k==0|| h==n+1|| k==m+1) Link ((I-1) *m+j+ -, +, INF); ElseLink (i-1) *m+j+ -, (H1) *m+K,inf); }} link ((I-1) *m+j, (I-1) *m+j+ -, mp[i][j]-'0'); }        }    }    intsum=0;  for(intI=1; i<=n; i++)    {         for(intj=1; j<=m; J + +)        {            if(mp2[i][j]=='L') {sum++; Link (999, (I-1) *m+j,1); } }} cout<<sum-dinic (999, +) <<Endl; return 0;}

"bzoj1066" [SCOI2007] Lizard network Max stream