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题解：“没有公共边”这是赤裸裸的二分图呀！随便按(i+j)&1造个二分图跑了就行。。。
而且，这是我第一个接触的构造题吧= = 还是非常有纪念意义的= =我记得当年还是小健建给我亲自敲了一遍DInic然后敲了一遍这道题。。。回忆满满呢~
现在老练的ISAP+缩行：
1 //这两道题有什么区别。。。= = 2 #include<iostream> 3 #include<cstdio> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #include<cstring> 8 using namespace std; 9 const int maxn=2002+10,maxm=80002+10,inf=-1u>>1; 10 struct ISAP{ 11 struct tedge{int x,y,w,next;}adj[maxm];int ms,fch[maxn]; 12 int d[maxn],s[maxn],cur[maxn],gap[maxn],n,top; 13 void init(int n){ 14 this->n=n;ms=0;top=0; 15 memset(d,-1,sizeof(d)); 16 memset(fch,-1,sizeof(fch)); 17 return; 18 } 19 void addedge(int u,int v,int w){ 20 adj[ms]=(tedge){u,v,w,fch[u]};fch[u]=ms++; 21 adj[ms]=(tedge){v,u,0,fch[v]};fch[v]=ms++; 22 return; 23 } 24 void bfs(){ 25 queue<int>Q;Q.push(n);d[n]=0; 26 while(!Q.empty()){ 27 int u=Q.front();Q.pop(); 28 for(int i=fch[u];i!=-1;i=adj[i].next){ 29 int v=adj[i].y; 30 if(d[v]==-1) d[v]=d[u]+1,Q.push(v); 31 } 32 } return; 33 } 34 int maxflow(int S,int T){ 35 n=T;bfs();int k=S,i,flow=0; 36 for(i=0;i<=n;i++) cur[i]=fch[i],gap[d[i]]++; 37 while(d[S]<n){ 38 if(k==n){ 39 int mi=inf,pos; 40 for(i=0;i<top;i++) if(adj[s[i]].w<mi) mi=adj[s[i]].w,pos=i; 41 for(i=0;i<top;i++) adj[s[i]].w-=mi,adj[s[i]^1].w+=mi; 42 flow+=mi;top=pos;k=adj[s[top]].x; 43 } 44 for(i=cur[k];i!=-1;i=adj[i].next){ 45 int v=adj[i].y; 46 if(adj[i].w&&d[k]==d[v]+1){cur[k]=i;k=v;s[top++]=i;break;} 47 } 48 if(i==-1){ 49 int lim=n; 50 for(i=fch[k];i!=-1;i=adj[i].next){ 51 int v=adj[i].y; 52 if(adj[i].w&&d[v]<lim) lim=d[v],cur[k]=i; 53 } if(--gap[d[k]]==0) break; 54 d[k]=lim+1;gap[d[k]]++; 55 if(k!=S) k=adj[s[--top]].x; 56 } 57 } return flow; 58 } 59 }sol; 60 inline int read(){ 61 int x=0,sig=1;char ch=getchar(); 62 while(!isdigit(ch)){if(ch==‘-‘) sig=-1;ch=getchar();} 63 while(isdigit(ch)) x=10*x+ch-‘0‘,ch=getchar(); 64 return x*=sig; 65 } 66 inline void write(int x){ 67 if(x==0){putchar(‘0‘);return;}if(x<0) putchar(‘-‘),x=-x; 68 int len=0,buf[15];while(x) buf[len++]=x%10,x/=10; 69 for(int i=len-1;i>=0;i--) putchar(buf[i]+‘0‘);return; 70 } 71 int A[21][21],mx[]={1,-1,0,0},my[]={0,0,-1,1}; 72 void init(){ 73 int n=read();long long ans=0; 74 sol.init(n*n+2);int S=n*n+1,T=n*n+2; 75 for(int i=1;i<=n;i++) 76 for(int j=1;j<=n;j++){ 77 A[i][j]=read(); 78 if((i+j)&1){ 79 sol.addedge(S,i*n+j-n,A[i][j]); 80 for(int d=0;d<4;d++){ 81 int nx=i+mx[d],ny=j+my[d]; 82 if(nx>=1&&nx<=n&&ny>=1&&ny<=n) sol.addedge(i*n+j-n,nx*n+ny-n,inf); 83 } 84 } 85 else sol.addedge(i*n+j-n,T,A[i][j]); 86 ans+=A[i][j]; 87 } 88 write(ans-sol.maxflow(S,T)); 89 return; 90 } 91 void work(){ 92 return; 93 } 94 void print(){ 95 return; 96 } 97 int main(){ 98 init();work();print();return 0; 99 }
当时最稚嫩的代码。。。真的有种说不出的感动：
1 #include <iostream> 2 #include <queue> 3 using namespace std; 4 5 const int maxn = 400 + 10; 6 const int maxm = 10000 + 10; 7 8 struct Edge 9 { 10 int from, to, cap, flow; 11 }; 12 13 struct Dinic 14 { 15 int n, m, s, t; 16 int first[maxn], next[maxm]; 17 18 Edge edges[maxm]; 19 20 void init(int n) 21 { 22 this -> n = n; 23 m = 0; 24 25 memset(first, -1, sizeof(first)); 26 27 return ; 28 } 29 30 void AddEdge(int from, int to, int cap) 31 { 32 edges[m] = (Edge){from, to, cap, 0}; 33 next[m] = first[from]; 34 first[from] = m++; 35 36 edges[m] = (Edge){to, from, 0, 0}; 37 next[m] = first[to]; 38 first[to] = m++; 39 40 return ; 41 } 42 43 int d[maxn], cur[maxn]; 44 bool vis[maxn]; 45 46 int BFS() 47 { 48 memset(vis, 0, sizeof(vis)); 49 queue<int> Q; 50 51 Q.push(s); 52 vis[s] = true; 53 d[s] = 0; 54 55 56 57 while(!Q.empty()) 58 { 59 int x = Q.front(); Q.pop(); 60 61 for(int i = first[x]; i != -1; i = next[i]) 62 { 63 Edge& e = edges[i]; 64 if(!vis[e.to] && e.cap > e.flow) 65 { 66 vis[e.to] = true; 67 d[e.to] = d[x] + 1; 68 Q.push(e.to); 69 } 70 } 71 } 72 73 return vis[t]; 74 } 75 int DFS(int x, int a) 76 { 77 if(x == t || !a) return a; 78 79 int f, flow = 0; 80 81 for(int& i = cur[x]; i != -1; i = next[i]) 82 { 83 Edge& e = edges[i]; 84 if(d[e.to] == d[x] + 1 && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) 85 { 86 flow += f; 87 a -= f; 88 e.flow += f; 89 edges[i ^ 1].flow -= f; 90 if(!a) break; 91 } 92 } 93 return flow; 94 } 95 96 97 int MaxFlow(int s, int t) 98 { 99 this -> s = s; 100 this -> t = t; 101 102 int flow = 0; 103 104 while(BFS()) 105 { 106 for(int i = 0; i < n; i++) cur[i] = first[i]; 107 flow += DFS(s, 1000000000); 108 } 109 return flow; //你大爷！！！！！！！！！！！！！！！！ 110 } 111 }sol; 112 113 int a[21][21]; 114 int nx[] = {0, 0, -1, 1}; 115 int ny[] = {-1, 1, 0, 0}; 116 117 int main() 118 { 119 int n, m; 120 scanf("%d", &n); 121 122 m = n; 123 124 sol.init(n * m + 2); 125 126 long long double tot = 0; 127 128 int s = n * m; 129 int t = n * m + 1; 130 131 for(int i = 0; i < n; i++) 132 for(int j = 0; j < m; j++) 133 { 134 scanf("%d", &a[i][j]); 135 tot += a[i][j]; 136 if((i + j) & 1) 137 { 138 sol.AddEdge(s, i * m + j, a[i][j]); 139 140 141 for(int d = 0; d < 4; d++) 142 { 143 int mx = nx[d] + i; 144 int my = ny[d] + j; 145 146 if(mx >= 0 && mx < n && my >= 0 && my < m) 147 { 148 sol.AddEdge(i * m + j, mx * m + my, 2000000000); 149 } 150 } 151 } 152 else sol.AddEdge(i * m + j, t, a[i][j]); 153 } 154 155 printf("%d\n", tot - sol.MaxFlow(s, t)); 156 157 //system("pause"); 158 return 0; 159 }
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原文地址：http://www.cnblogs.com/chxer/p/4542665.html
