Hướng dẫn giải của Trồng cây
Chỉ dùng lời giải này khi không có ý tưởng, và đừng copy-paste code từ lời giải này. Hãy tôn trọng người ra đề và người viết lời giải.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này
Code mẫu của flashmt
const max=250; var a,b:array[1..max,1..max] of byte; h,v:array[1..max] of byte; m,n:byte; dem:longint; procedure rf; var i,j:byte; begin fillchar(h,sizeof(h),0); fillchar(v,sizeof(v),0); readln(m,n); for i:=1 to m do begin for j:=1 to n do begin read(a[i,j]); if a[i,j]=1 then begin inc(h[i]); inc(v[j]); end; end; readln; end; end; function find(var i:byte):byte; var j:byte; begin for j:=1 to m do if h[j] mod 2 = 1 then begin i:=j; exit(1); end; for j:=1 to n do if v[j] mod 2 = 1 then begin i:=j; exit(2); end; for j:=1 to m do if h[j]>0 then begin i:=j; exit(3); end; for j:=1 to n do if v[j]>0 then begin i:=j; exit(4); end; exit(0); end; procedure work(k,c:byte); var i:byte; begin if (k=1) or (k=3) then begin repeat for i:=1 to n do if a[c,i]=1 then begin b[c,i]:=1+dem mod 2; inc(dem); dec(h[c]); dec(v[i]); a[c,i]:=0; break; end; c:=i; if v[c]=0 then break; for i:=1 to m do if a[i,c]=1 then begin b[i,c]:=1+dem mod 2; inc(dem); dec(h[i]); dec(v[c]); a[i,c]:=0; break; end; c:=i; if h[c]=0 then break; until false; end else begin repeat for i:=1 to m do if a[i,c]=1 then begin b[i,c]:=1+dem mod 2; inc(dem); dec(h[i]); dec(v[c]); a[i,c]:=0; break; end; c:=i; if h[c]=0 then break; for i:=1 to n do if a[c,i]=1 then begin b[c,i]:=1+dem mod 2; inc(dem); dec(h[c]); dec(v[i]); a[c,i]:=0; break; end; c:=i; if v[c]=0 then break; until false; end; end; procedure pr; var i,j:byte; begin fillchar(b,sizeof(b),0); dem:=1; repeat i:=find(j); if i=0 then break; work(i,j); until false; end; procedure wf; var i,j:byte; begin for i:=1 to m do begin for j:=1 to n do write(b[i,j],' '); writeln; end; end; begin rf; pr; wf; end.
Code mẫu của RR
#include <iostream> #include <algorithm> #include <vector> #define FOR(i,a,b) for(int i=a; i<=b; i++) #define FORD(i,a,b) for(int i=a; i>=b; i--) #define FORV(i,v) for(typeof(v.begin()) i=v.begin(); i!=v.end(); i++) #define MAXN 255 #define P pair<int,int> #define X first #define Y second #define PB push_back using namespace std; int m,n,a[MAXN][MAXN],ind[MAXN][MAXN],npoint,color[MAXN*MAXN]; P pos[MAXN*MAXN]; vector<int> ke[MAXN*MAXN]; void inp() { scanf("%d %d",&m,&n); FOR(i,1,m) FOR(j,1,n) { scanf("%d",&a[i][j]); if (a[i][j]) { npoint++; pos[npoint].X=i; pos[npoint].Y=j; ind[i][j]=npoint; } } } void init() { int last; FOR(i,1,m) { int cnt=0; FOR(j,1,n) if (a[i][j]) { if (cnt&1) { ke[ind[i][j]].PB(last); ke[last].PB(ind[i][j]); } cnt++; last=ind[i][j]; } } FOR(j,1,n) { int cnt=0; FOR(i,1,m) if (a[i][j]) { if (cnt&1) { ke[ind[i][j]].PB(last); ke[last].PB(ind[i][j]); } cnt++; last=ind[i][j]; } } } int first,last,qu[MAXN*MAXN],n1,n2; inline void paint(int u,int k) { color[u]=k; if (k==1) n1++; else n2++; } void bfs(int u) { first=1; last=1; qu[1]=u; if (n1<n2) paint(u,1); else paint(u,2); while (first<=last) { u=qu[first]; first++; FORV(v,ke[u]) if (!color[*v]) { paint(*v,3-color[u]); last++; qu[last]=*v; } } } void solve() { n1 = n2 = 0; FOR(i,1,m) FOR(j,1,n) if (a[i][j] && !color[ind[i][j]]) { bfs(ind[i][j]); } FOR(i,1,m) { FOR(j,1,n) if (!a[i][j]) printf("0 "); else printf("%d ",color[ind[i][j]]); printf("\n"); } } int main() { // freopen("input.txt","r",stdin); // freopen("output.txt","w",stdout); inp(); init(); solve(); return 0; }
Code mẫu của hieult
#include <set> #include <map> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <string> #include <vector> #include <cstdlib> #include <cstring> #include <sstream> #include <iomanip> #include <iostream> #include <algorithm> #include <ctime> #include <deque> #include <bitset> #include <cctype> #include <utility> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned int ui; typedef unsigned long long ull; #define Rep(i,n) for(int i = 0; i < (n); ++i) #define Repd(i,n) for(int i = (n)-1; i >= 0; --i) #define For(i,a,b) for(int i = (a); i <= (b); ++i) #define Ford(i,a,b) for(int i = (a); i >= (b); --i) #define Fit(i,v) for(__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) #define Fitd(i,v) for(__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++i) #define mp make_pair #define pb push_back #define fi first #define se second #define sz(a) ((int)(a).size()) #define all(a) (a).begin(), (a).end() #define ms(a,x) memset(a, x, sizeof(a)) template<class F, class T> T convert(F a, int p = -1) { stringstream ss; if (p >= 0) ss << fixed << setprecision(p); ss << a; T r; ss >> r; return r; } template<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; } template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<class T> T sqr(T x) { return x * x; } template<class T> T cube(T x) { return x * x * x; } template<class T> int getbit(T s, int i) { return (s >> i) & 1; } template<class T> T onbit(T s, int i) { return s | (T(1) << i); } template<class T> T offbit(T s, int i) { return s & (~(T(1) << i)); } template<class T> int cntbit(T s) { return s == 0 ? 0 : cntbit(s >> 1) + (s & 1); } const int bfsz = 1 << 16; char bf[bfsz + 5]; int rsz = 0; int ptr = 0; char gc() { if (rsz <= 0) { ptr = 0; rsz = (int) fread(bf, 1, bfsz, stdin); if (rsz <= 0) return EOF; } --rsz; return bf[ptr++]; } void ga(char &c) { c = EOF; while (!isalpha(c)) c = gc(); } int gs(char s[]) { int l = 0; char c = gc(); while (isspace(c)) c = gc(); while (c != EOF && !isspace(c)) { s[l++] = c; c = gc(); } s[l] = '\0'; return l; } template<class T> bool gi(T &v) { v = 0; char c = gc(); while (c != EOF && c != '-' && !isdigit(c)) c = gc(); if (c == EOF) return false; bool neg = c == '-'; if (neg) c = gc(); while (isdigit(c)) { v = v * 10 + c - '0'; c = gc(); } if (neg) v = -v; return true; } typedef pair<int, int> II; const ld PI = acos(-1.0); const ld eps = 1e-12; const int dr[] = { -1, 0, +1, 0}; const int dc[] = { 0, +1, 0, -1}; const int inf = (int) 1e9 + 5; const ll linf = (ll) 1e18 + 5; const int mod = (ll) (1e9 + eps); using namespace std; #define maxv 1005 #define maxe 250005 struct Dinic { int n, s, t, E, adj[maxe], flow[maxe], cap[maxe], next[maxe], last[maxv], run[maxv], level[maxv], que[maxv]; void init(int _n, int _s, int _t) { n = _n; s = _s; t = _t; E = 0; For(i, 0, n) last[i] = -1; } void add(int u, int v, int c1, int c2) { adj[E] = v; flow[E] = 0; cap[E] = c1; next[E] = last[u]; last[u] = E++; adj[E] = u; flow[E] = 0; cap[E] = c2; next[E] = last[v]; last[v] = E++; } bool bfs() { For(i, 0, n) level[i] = -1; level[s] = 0; int qsize = 0; que[qsize++] = s; Rep(i, qsize) { for (int u = que[i], e = last[u]; e != -1; e = next[e]) { int v = adj[e]; if (flow[e] < cap[e] && level[v] == -1) { level[v] = level[u] + 1; que[qsize++] = v; } } } return level[t] != -1; } int dfs(int u, int bot) { if (u == t) return bot; for (int &e = run[u]; e != -1; e = next[e]) { int v = adj[e], delta = 0; if (level[v] == level[u] + 1 && flow[e] < cap[e] && (delta = dfs(v, min(bot, cap[e] - flow[e]))) > 0) { flow[e] += delta; flow[e ^ 1] -= delta; return delta; } } return 0; } int maxflow() { int total = 0; while (bfs()) { For(i, 0, n) run[i] = last[i]; for (int delta = dfs(s, inf); delta > 0; delta = dfs(s, inf)) total += delta; } return total; } } dinic; int n, m, row[maxv], col[maxv], oddrow, oddcol, a[maxv][maxv], sum; int main(){ // freopen("in.txt", "r", stdin); cin >> n >> m; ms(row, 0); ms(col, 0); sum = 0; dinic.init(n + m + 4, n + m + 3, n + m + 4); For(i, 1, n) For(j, 1, m) { scanf("%d", &a[i][j]); row[i] += a[i][j]; col[j] += a[i][j]; sum += a[i][j]; } oddrow = 0; oddcol = 0; For(i, 1, n) oddrow += (row[i] & 1); For(i, 1, m) oddcol += (col[i] & 1); For(i, 1, n) For(j, 1, m) if(a[i][j]){ dinic.add(i, j + n, 1, 0); } For(i, 1, n) dinic.add(n + m + 1, i, row[i] / 2, 0); For(i, 1, m) dinic.add(i + n, n + m + 2, col[i] / 2, 0); dinic.add(n + m + 3, n + m + 1, sum / 2, 0); dinic.add(n + m + 2, n + m + 4, sum / 2, 0); dinic.maxflow(); if(oddrow > oddcol){ Rep(i, n) dinic.cap[i * 2 + sum * 2] = row[i + 1] - row[i + 1] / 2; dinic.maxflow(); Rep(i, m) dinic.cap[i * 2 + n * 2 + sum * 2] = col[i + 1] - col[i + 1] / 2; dinic.maxflow(); } else{ Rep(i, m) dinic.cap[i * 2 + n * 2 + sum * 2] = col[i + 1] - col[i + 1] / 2; dinic.maxflow(); Rep(i, n) dinic.cap[i * 2 + sum * 2] = row[i + 1] - row[i + 1] / 2; dinic.maxflow(); } int run = 0; For(i, 1, n) For(j, 1, m){ if(a[i][j]){ if(dinic.flow[run]) printf("1%c", j == m ? '\n' : ' '); else printf("2%c", j == m ? '\n' : ' '); run += 2; } else printf("0%c", j == m ? '\n' : ' '); } return 0; }
Code mẫu của ll931110
{$R+,Q+} {$MODE DELPHI} program GARDEN25; const input = ''; output = ''; maxn = 250; var hz,ve,a,res: array[1..maxn,1..maxn] of integer; free: array[1..maxn,1..maxn] of boolean; m,n,delta: integer; procedure init; var f: text; i,j: integer; begin assign(f, input); reset(f); readln(f, m, n); for i := 1 to m do for j := 1 to n do read(f, a[i,j]); close(f); delta := 0; end; procedure construct; var t,i,j: integer; begin fillchar(hz, sizeof(hz), 0); fillchar(ve, sizeof(ve), 0); for i := 1 to m do begin t := 0; for j := 1 to n do if a[i,j] = 1 then if t = 0 then t := j else begin hz[i,t] := j; hz[i,j] := t; t := 0; end; end; for j := 1 to n do begin t := 0; for i := 1 to m do if a[i,j] = 1 then if t = 0 then t := i else begin ve[t,j] := i; ve[i,j] := t; t := 0; end; end; end; procedure DFS(i,j: integer); var k,t: integer; begin free[i,j] := false; t := hz[i,j]; if (t <> 0) and (a[i,t] = 1) then if free[i,t] then begin res[i,t] := 3 - res[i,j]; if res[i,t] = 1 then inc(delta) else dec(delta); DFS(i,t); end; t := ve[i,j]; if (t <> 0) and (a[t,j] = 1) then if free[t,j] then begin res[t,j] := 3 - res[i,j]; if res[t,j] = 1 then inc(delta) else dec(delta); DFS(t,j); end; end; procedure solve; var i,j: integer; begin fillchar(free, sizeof(free), true); fillchar(res, sizeof(res), 0); for i := 1 to m do for j := 1 to n do if free[i,j] and (a[i,j] = 1) then begin if delta <= -1 then begin inc(delta); res[i,j] := 1; end else begin dec(delta); res[i,j] := 2; end; DFS(i,j); end; end; procedure printresult; var f: text; i,j: integer; begin assign(f, output); rewrite(f); for i := 1 to m do begin write(f, res[i,1]); for j := 2 to n do write(f, ' ', res[i,j]); if i <> m then writeln(f); end; close(f); end; begin init; construct; solve; printresult; end.
Code mẫu của skyvn97
#include<bits/stdc++.h> #define MAX 555 #define FOR(i,a,b) for (int i=(a);i<=(b);i=i+1) #define REP(i,n) for (int i=0;i<(n);i=i+1) #define FORE(i,v) for (__typeof((v).begin()) i=(v).begin();i!=(v).end();i++) using namespace std; const int INF=(int)1e9+7; class DinicFlow { public: vector<int> dist,head,q,work; vector<int> point,capa,flow,next; int n,m; DinicFlow() { n=0;m=0; } DinicFlow(int n) { this->n=n;m=0; dist.assign(n+7,0); head.assign(n+7,-1); q.assign(n+7,0); work.assign(n+7,0); } void addedge(int u,int v,int c1,int c2) { point.push_back(v);capa.push_back(c1);flow.push_back(0);next.push_back(head[u]);head[u]=m++; point.push_back(u);capa.push_back(c2);flow.push_back(0);next.push_back(head[v]);head[v]=m++; } bool bfs(int s,int t) { FOR(i,1,n) dist[i]=-1; int sz=0; q[sz++]=s;dist[s]=0; REP(x,sz) { int u=q[x]; for (int i=head[u];i>=0;i=next[i]) if (dist[point[i]]<0 && flow[i]<capa[i]) { dist[point[i]]=dist[u]+1; q[sz++]=point[i]; } } return (dist[t]>=0); } int dfs(int s,int t,int f) { if (s==t) return (f); for (int &i=work[s];i>=0;i=next[i]) if (dist[point[i]]==dist[s]+1 && flow[i]<capa[i]) { int d=dfs(point[i],t,min(f,capa[i]-flow[i])); if (d>0) { flow[i]+=d; flow[i^1]-=d; return (d); } } return (0); } int getflow(int s,int t) { int totflow=0; while (bfs(s,t)) { FOR(i,1,n) work[i]=head[i]; while (true) { int d=dfs(s,t,INF); if (d<=0) break; totflow+=d; } } return (totflow); } }; class DemandFlow { public: DinicFlow G; vector<int> si,so; int n; DemandFlow() { n=0; } DemandFlow(int n) { this->n=n; si.assign(n+7,0); so.assign(n+7,0); G=DinicFlow(n+2); } void addedge(int u,int v,int d,int c) { si[v]+=d; so[u]+=d; G.addedge(u,v,c-d,0); } bool haveflow(int s,int t) { int sd=0; FOR(i,1,n) { sd+=si[i]; if (si[i]>0) G.addedge(n+1,i,si[i],0); if (so[i]>0) G.addedge(i,n+2,so[i],0); } G.addedge(t,s,INF-7,0); return (G.getflow(n+1,n+2)==sd); } }; DemandFlow DF; int n,m,cnt; int sr[MAX],sc[MAX],a[MAX][MAX]; void process(void) { scanf("%d%d",&m,&n); DF=DemandFlow(m+n+3); FOR(i,1,m) FOR(j,1,n) { scanf("%d",&a[i][j]); if (a[i][j]) { sr[i]++; sc[j]++; cnt++; DF.addedge(i,m+j,0,1); } } FOR(i,1,m) DF.addedge(m+n+1,i,sr[i]/2,sr[i]-sr[i]/2); FOR(i,1,n) DF.addedge(m+i,m+n+2,sc[i]/2,sc[i]-sc[i]/2); DF.addedge(m+n+3,m+n+1,cnt/2,cnt-cnt/2); assert(DF.haveflow(m+n+3,m+n+2)); int id=0; FOR(i,1,m) FOR(j,1,n) { if (a[i][j]) printf("%d",2-DF.G.flow[2*id++]); else printf("0"); if (j<n) printf(" "); else printf("\n"); } } int main(void) { #ifndef ONLINE_JUDGE freopen("tmp.txt","r",stdin); #endif int id=0; process(); return 0; }
Code mẫu của khuc_tuan
#include <iostream> #include <cstdio> using namespace std; int m, n; int a[255][255], b[255][255]; int degx[255], degy[255]; // int start_color; int C[3]; void gox(int i, int color); void goy(int j, int color); void gox(int i, int color) { for(int j=0;j<n;++j) if(a[i][j] == 1 && b[i][j] == 0) { b[i][j] = color; ++C[color]; --degx[i]; --degy[j]; goy( j, 3 - color); return; } } void goy(int j, int color) { for(int i=0;i<m;++i) if(a[i][j] == 1 && b[i][j] == 0) { b[i][j] = color; ++C[color]; --degx[i]; --degy[j]; gox( i, 3 - color); return; } } int main() { scanf("%d%d", &m, &n); for(int i=0;i<m;++i) for(int j=0;j<n;++j) scanf("%d", &a[i][j]); for(int i=0;i<m;++i) for(int j=0;j<n;++j) if(a[i][j]) { ++degx[i]; ++degy[j]; } // start_color = 1; for(int i=0;i<m;++i) if(degx[i] % 2 == 1) { if(C[1] > C[2]) gox( i, 2); else gox( i, 1); // start_color = 3 - start_color; } for(int j=0;j<n;++j) if(degy[j] % 2 == 1) { if(C[1] > C[2]) goy( j, 2); else goy( j, 1); // goy( j, start_color); // start_color = 3 - start_color; } for(int i=0;i<m;++i) if(degx[i] > 0) gox( i, 1); for(int j=0;j<n;++j) if(degy[j] > 0) goy( j, 1); for(int i=0;i<m;++i) { for(int j=0;j<n;++j) printf("%d ", b[i][j]); printf("\n"); } // system("pause"); return 0; }
Bình luận