結果
問題 | No.421 しろくろチョコレート |
ユーザー |
|
提出日時 | 2016-09-10 00:27:18 |
言語 | C++11 (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 6 ms / 2,000 ms |
コード長 | 4,256 bytes |
コンパイル時間 | 1,118 ms |
コンパイル使用メモリ | 107,136 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-09-23 07:07:48 |
合計ジャッジ時間 | 2,634 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
other | AC * 65 |
コンパイルメッセージ
main.cpp: In member function ‘void Dinic::add_edge(int, int, int)’: main.cpp:118:62: warning: narrowing conversion of ‘(&((Dinic*)this)->Dinic::graph.std::vector<std::vector<Dinic::edge> >::operator[](((std::vector<std::vector<Dinic::edge> >::size_type)to)))->std::vector<Dinic::edge>::size()’ from ‘std::vector<Dinic::edge>::size_type’ {aka ‘long unsigned int’} to ‘int’ [-Wnarrowing] 118 | graph[from].push_back((edge) {to, cap, graph[to].size()}); | ~~~~~~~~~~~~~~^~ main.cpp:119:65: warning: narrowing conversion of ‘((&((Dinic*)this)->Dinic::graph.std::vector<std::vector<Dinic::edge> >::operator[](((std::vector<std::vector<Dinic::edge> >::size_type)from)))->std::vector<Dinic::edge>::size() - 1)’ from ‘std::vector<Dinic::edge>::size_type’ {aka ‘long unsigned int’} to ‘int’ [-Wnarrowing] 119 | graph[to].push_back((edge) {from, 0, graph[from].size() - 1}); | ~~~~~~~~~~~~~~~~~~~^~~
ソースコード
#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iomanip>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <utility>#include <vector>#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)using namespace std;typedef long long int ll;typedef vector<int> VI;typedef vector<ll> VL;typedef pair<int, int> PI;const ll mod = 1e9 + 7;const int DEBUG = 1;const int N = 51;string s[N];int n, m;pair<int, VL> check(int row) {int k = 0;VL t;int cur = 0;int cnt = 0;REP(i, 0, m + 1) {if (i < m && s[row][i] != '.') {if (cnt == 0) {cur = i;}cnt++;} else {k += cnt / 2;if (cnt % 2 != 0) {ll acc = 0;REP(j, 0, cnt / 2 + 1) {acc |= 1LL << (cur + 2 * j);}t.push_back(acc);}cnt = 0;}}return pair<int, VL>(k, t);}/*** Dinic's algorithm for maximum flow problem.* Header requirement: vector, queue* Verified by: ABC010-D(http://abc010.contest.atcoder.jp/submissions/602810)*/class Dinic {private:struct edge {int to, cap, rev; // rev is the position of reverse edge in graph[to]};std::vector<std::vector<edge> > graph;std::vector<int> level;std::vector<int> iter;/* Perform bfs and calculate distance from s */void bfs(int s) {level.assign(level.size(), -1);std::queue<int> que;level[s] = 0;que.push(s);while (! que.empty()) {int v = que.front(); que.pop();for (int i = 0; i < graph[v].size(); ++i) {edge &e = graph[v][i];if (e.cap > 0 && level[e.to] == -1) {level[e.to] = level[v] + 1;que.push(e.to);}}}}/* search augment path by dfs.if f == -1, f is treated as infinity. */int dfs(int v, int t, int f) {if (v == t) {return f;}for (int &i = iter[v]; i < graph[v].size(); ++i) {edge &e = graph[v][i];if (e.cap > 0 && level[v] < level[e.to]) {int newf = f == -1 ? e.cap : std::min(f, e.cap);int d = dfs(e.to, t, newf);if (d > 0) {e.cap -= d;graph[e.to][e.rev].cap += d;return d;}}}return 0;}public:/* v is the number of vertices (labeled from 0 .. v-1) */Dinic(int v) : graph(v), level(v, -1), iter(v, 0) {}void add_edge(int from, int to, int cap) {graph[from].push_back((edge) {to, cap, graph[to].size()});graph[to].push_back((edge) {from, 0, graph[from].size() - 1});}int max_flow(int s, int t) {int flow = 0;while (1) {bfs(s);if (level[t] < 0) {return flow;}iter.assign(iter.size(), 0);int f;while ((f = dfs(s, t, -1)) > 0) {flow += f;}}}};int calc(void) {Dinic din(n * m + 2);REP(i, 0, n) {REP(j, 0, m) {if (s[i][j] == '.') continue;int dxy[5] = {1, 0, -1, 0, 1};REP(d, 0, 4) {int nx = i + dxy[d];int ny = j + dxy[d + 1];if (nx < 0 || nx >= n || ny < 0 || ny >= m) {continue;}if (s[nx][ny] != '.') {if ((i + j) % 2) {din.add_edge(i * m + j, nx * m + ny, 1);} else {din.add_edge(nx * m + ny, i * m + j, 1);}}}if ((i + j) % 2) {din.add_edge(n * m, i * m + j, 1);} else {din.add_edge(i * m + j, n * m + 1, 1);}}}return din.max_flow(n * m, n * m + 1);}int main(void){cin >> n >> m;REP(i, 0, n) {cin >> s[i];}int w = 0;int b = 0;REP(i, 0, n) {REP(j, 0, m) {if (s[i][j] == 'w') w++;if (s[i][j] == 'b') b++;}}int c = calc();w -= c;b -= c;int mi = min(w, b);cout << c * 100 + mi * 8 + w + b << endl;}