首先,这种相邻格子的问题都会联系到二分图:横纵坐标和为偶数的在左边,横纵坐标和为奇数的在右边。
构图如下:
原点和左边的点相连接,容量是其权值。右边的点和汇点连接,容量是其权值(如果某点为必选的,则权值为INF)。如果左边的点x和右边的点y相邻,连接x,y容量为2 * (x & y)。
每个割都代表一种方案,最小化代价则会得到最大价值。
#include <cstdio> #include <cstring> #include <vector> #include <queue> using namespace std; #define maxn 3000 #define INF 100000 struct Edge { int from, to, cap, flow; }; int n, m, s, t; int sum[55][55]; int mark[55][55]; int Count[55][55]; vector<Edge> edges; // 边数的两倍 vector<int> G[maxn]; // 邻接表,G[i][j]表示结点i的第j条边在e数组中的序号 bool vis[maxn]; // BFS使用 int d[maxn]; // 从起点到i的距离 int cur[maxn]; // 当前弧指针 int min(int a,int b) { if(a<b) return a; else return b; } void AddEdge(int from, int to, int cap) { int len; Edge temp; temp.from=from;temp.to=to;temp.cap=cap;temp.flow=0; edges.push_back(temp); temp.from=to;temp.to=from;temp.cap=0;temp.flow=0; edges.push_back(temp); len = edges.size(); G[from].push_back(len-2); G[to].push_back(len-1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue<int> Q; Q.push(s); vis[s] = 1; d[s] = 0; 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) { e.flow += f; edges[G[x][i]^1].flow -= f; flow += f; a -= f; if(a == 0) break; } } if(flow==0) d[x]=-1; return flow; } int Maxflow() { int flow = 0; while(BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } int main() { int k; int i,j; int x,y; while(scanf("%d%d%d",&n,&m,&k)!=EOF) { int now=0; int ans=0; memset(mark,0,sizeof(mark)); s=0;t=n*m+1; for(i=1;i<=n;i++) for(j=1;j<=m;j++) Count[i][j]=++now; for(i=0;i<=t;i++) G[i].clear(); for(i=1;i<=n;i++) for(j=1;j<=m;j++) { scanf("%d",&sum[i][j]); ans+=sum[i][j]; } for(i=1;i<=k;i++) { scanf("%d%d",&x,&y); mark[x][y]=1; } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if((i+j)%2==0) { if(mark[i][j]==0) AddEdge(s,Count[i][j],sum[i][j]); if(mark[i][j]==1) AddEdge(s,Count[i][j],INF); if(i-1>0) AddEdge(Count[i][j],Count[i-1][j],2*(sum[i][j]&sum[i-1][j])); if(i+1<=n) AddEdge(Count[i][j],Count[i+1][j],2*(sum[i][j]&sum[i+1][j])); if(j-1>0) AddEdge(Count[i][j],Count[i][j-1],2*(sum[i][j]&sum[i][j-1])); if(j+1<=m) AddEdge(Count[i][j],Count[i][j+1],2*(sum[i][j]&sum[i][j+1])); } else { if(mark[i][j]==0) AddEdge(Count[i][j],t,sum[i][j]); else AddEdge(Count[i][j],t,INF); } } } ans=ans-Maxflow(); printf("%d ",ans); } return 0; }