网络最大流
TLE了两天的题目。80次Submit才AC,发现是刘汝佳白书的Dinic代码还可以优化。。。。。瞬间无语。。。。。
#include<cstdio> #include<cstring> #include<string> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; const int maxn = 30000 + 10; const int INF = 0x7FFFFFFF; struct Edge { int from, to, cap, flow; Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {} }; vector<Edge>edges; vector<int>G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; int n, m, s, t; void init() { for (int i = 0; i < maxn; i++) G[i].clear(); edges.clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); int w = edges.size(); G[from].push_back(w - 2); G[to].push_back(w - 1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue<int>Q; Q.push(s); d[s] = 0; vis[s] = 1; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (!vis[e.to] && e.cap>e.flow) { vis[e.to] = 1; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int DFS(int x, int a) { if (x == t || a == 0) return a; int flow = 0, f; for (int &i = cur[x]; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0) { edges[G[x][i]].flow+=f; edges[G[x][i] ^ 1].flow-=f; flow+=f; a-=f; if(a==0) break; } } if(!flow) d[x] = -1; return flow; } int dinic(int s, int t) { int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } //输入输出 int N, M; char S[1000]; int Map1[105][105];//横向 int Map2[105][105];//纵向 int BH[105][105]; int FLAG[maxn]; int main() { while (~scanf("%d%d", &N, &M)) { edges.clear(); for (int i = 0; i<maxn; i++) G[i].clear(); memset(Map1, 0, sizeof(Map1)); memset(Map2, 0, sizeof(Map2)); memset(BH, 0, sizeof(BH)); for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++) { scanf("%s", S); if (S[3] == '.') continue; if (S[3] == 'X') Map1[i][j] = -1, Map2[i][j] = -1; else { if (S[2] == 'X') { Map2[i][j] = -1; Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1; } else if (S[4] == 'X') { Map1[i][j] = -1; Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1; } else { Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1; Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1; } } } int Tot = 1; for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++) if (Map1[i][j] == 0) BH[i][j] = Tot, Tot++; int Duan = Tot - 1;//方格编号1--Duan s = 0; t = 30000; for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++) { if (Map1[i][j] != -1 && Map1[i][j] != 0) { int Zong = 0; for (int k = j + 1;; k++) { if (k>M || Map1[i][k] != 0) break; AddEdge(Tot, BH[i][k], 8); Zong++; } AddEdge(s, Tot, Map1[i][j] - Zong); Tot++; } } for (int i = 1; i <= N; i++) for (int j = 1; j <= M; j++) { if (Map2[i][j] != -1 && Map2[i][j] != 0) { int Zong = 0; for (int k = i + 1;; k++) { if (Map2[k][j] != 0 || k>N) break; AddEdge(BH[k][j], Tot, 8); Zong++; } AddEdge(Tot, t, Map2[i][j] - Zong); Tot++; } } dinic(s, t); memset(FLAG, 0, sizeof FLAG); for (int i = 0; i<edges.size(); i = i + 2) if (edges[i].from >= 1 && edges[i].from <= Duan) FLAG[edges[i].from] = edges[i].flow; for (int i = 1; i <= N; i++) { for (int j = 1; j <= M; j++) { if (BH[i][j] == 0) printf("_"); else printf("%d", FLAG[BH[i][j]] + 1); if (j<M) printf(" "); } printf(" "); } } return 0; }