結果
問題 | No.421 しろくろチョコレート |
ユーザー |
|
提出日時 | 2019-03-27 23:03:55 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 4,178 bytes |
コンパイル時間 | 2,100 ms |
コンパイル使用メモリ | 187,016 KB |
実行使用メモリ | 101,888 KB |
最終ジャッジ日時 | 2024-10-11 06:16:03 |
合計ジャッジ時間 | 12,482 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 18 WA * 47 |
ソースコード
#include <bits/stdc++.h> #define FOR(v, a, b) for(int v = (a); v < (b); ++v) #define FORE(v, a, b) for(int v = (a); v <= (b); ++v) #define REP(v, n) FOR(v, 0, n) #define REPE(v, n) FORE(v, 0, n) #define REV(v, a, b) for(int v = (a); v >= (b); --v) #define ALL(x) (x).begin(), (x).end() #define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it) #define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it) #define EXIST(c,x) ((c).find(x) != (c).end()) #define LLI long long int #define fst first #define snd second #ifdef DEBUG #include <misc/C++/Debug.cpp> #else #define dump(x) ((void)0) #endif using namespace std; #define gcd __gcd template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;} template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;} template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;} template <typename T, typename U> istream& operator>>(istream &is, pair<T,U> &p){is >> p.first >> p.second; return is;} template <typename T, typename U> T& chmin(T &a, const U &b){return a = (a<=b?a:b);} template <typename T, typename U> T& chmax(T &a, const U &b){return a = (a>=b?a:b);} template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);} template <typename T, T INF> class Dinic{ private: vector<vector<pair<int,T>>> graph; int size, s, t; vector<vector<T>> cap; vector<int> level; bool buildLevel(){ fill(ALL(level), 0); level[s] = 1; deque<int> deq = {s}; while(!deq.empty()){ int cur = deq.front(); deq.pop_front(); REP(i,size) if(level[i]==0 && cap[cur][i]>0){ level[i] = level[cur] + 1; deq.push_back(i); } } return level[t] != 0; } void dfs(vector<int> &path, T &flow){ if(path.empty()) return; int cur = path.back(); if(cur == t){ T f = INF; FOR(i,1,path.size()) f = min(f, cap[path[i-1]][path[i]]); FOR(i,1,path.size()){ cap[path[i-1]][path[i]] -= f; cap[path[i]][path[i-1]] += f; } flow += f; }else{ REP(i,size){ if(cap[cur][i]>0 && level[i]>level[cur]){ path.push_back(i); dfs(path, flow); path.pop_back(); } } } } T augment(){ T f = 0; vector<int> path = {s}; dfs(path, f); return f; } T loop(){ T f = 0; while(buildLevel()) f += augment(); return f; } public: Dinic(vector<vector<pair<int,T>>> &_graph): graph(_graph), size(graph.size()) {} Dinic(int size): graph(size), size(size){} void add_edge(int from, int to, const T &cap){ graph[from].push_back({to, cap}); } T max_flow(int _s, int _t){ cap = vector<vector<T>>(size, vector<T>(size, 0)); level = vector<int>(size, 0); REP(i,size) for(auto &p : graph[i]){ int j = p.first; T d = p.second; cap[i][j] += d; } s = _s; t = _t; return loop(); } }; class BipartiteMatching{ public: int x, y; Dinic<int,INT_MAX> mflow; int s, t; BipartiteMatching(int x, int y): x(x), y(y), mflow(x+y+2), s(x+y), t(s+1){ REP(i,x) mflow.add_edge(s,i,1); REP(i,y) mflow.add_edge(x+i,t,1); } void add_edge(int i, int j){ mflow.add_edge(i,x+j,1); } int matching(){ return mflow.max_flow(s,t); } }; const int dir4[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; int main(){ cin.tie(0); ios::sync_with_stdio(false); int N,M; while(cin >> N >> M){ vector<string> S(N); cin >> S; int c = 0; REP(i,N) REP(j,M) if(S[i][j] != '.') ++c; BipartiteMatching bm(N*M,N*M); vector<vector<int>> w(N, vector<int>(M,-1)), b(N, vector<int>(M,-1)); int cw=0, cb=0; REP(i,N) REP(j,M){ if(S[i][j] == 'w') w[i][j] = cw++; if(S[i][j] == 'b') b[i][j] = cb++; } REP(i,N){ REP(j,M){ for(auto &d : dir4){ int y=i+d[0], x=j+d[1]; if(y<0 or y>=N or x<0 or x>=M) continue; if(S[i][j] == 'w' and S[y][x] == 'b') bm.add_edge(w[i][j], b[y][x]); } } } int mat = bm.matching(); int ans = mat*100 + (c-mat*2)/2*10 + c%2; cout << ans << endl; } return 0; }