Drainage Ditches（最大流入门）
作者：互联网
EK+邻接矩阵：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 #define rep(i,first,last) for(int i=first;i<=last;i++)
5 #define dep(i,first,last) for(int i=first;i>=last;i--)
6 const int maxn=205;
7 const int inf=0x3f3f3f3f;
8 int graph[maxn][maxn];
9 int pre[maxn];//记录前驱节点
10 int vis[maxn];
11 int n,m,a,b,flow;
12
13 int BFS(int st,int ed){
14 memset(vis,0,sizeof(vis));
15 queue<int>que;
16 while(!que.empty()) que.pop();
17 que.push(st);
18 vis[st]=1;
19 int curr,flow=inf;
20 pre[st]=pre[ed]=-1;
21 while(!que.empty()){
22 curr=que.front();
23 que.pop();
24 if( curr==ed ) break;
25 rep(nxt,1,n){
26 if(!vis[nxt]&&graph[curr][nxt]!=0){
27 vis[nxt]=1;
28 if( flow>graph[curr][nxt]) flow=graph[curr][nxt];
29 pre[nxt]=curr;
30 que.push(nxt);
31 }
32 }
33 }
34 if(pre[ed]==-1) return -1;//没有找到增广路
35 return flow;
36 }
37
38 int EK(int st,int ed){
39 int stream;
40 int sum=0;
41 while((stream=BFS(st,ed))!=-1 ){ //找到了增广路
42 int u=ed;
43 while( pre[u]!=-1 ){
44 graph[pre[u]][u]-=stream;//改变正向边的容量
45 graph[u][pre[u]]+=stream;//改变反向边的容量
46 u=pre[u];
47 }
48 sum += stream;
49 }
50 return sum;
51 }
52 int main()
53 {
54 while(~scanf("%d%d",&m,&n)){
55 memset(graph,0,sizeof(graph));
56 memset(vis,0,sizeof(vis));
57 while( m-- ){
58 scanf("%d%d%d",&a,&b,&flow);
59 graph[a][b]+=flow;//此处注意可能出现多条同一起点终点的情况
60 }
61 printf("%d\n",EK(1,n));
62 }
63 return 0;
64 }
View Code
EK+前向星：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int head[maxn<<1],cnt;
9 struct node{int to,nxt,val;}e[maxn<<1];
10 void addedge(int u,int v,int val){
11 e[cnt].to=v;
12 e[cnt].val=val;
13 e[cnt].nxt=head[u];
14 head[u]=cnt++;
15 }
16 int vis[maxn],pre[maxn];
17 int BFS(int st,int ed){
18 memset(vis,0,sizeof(vis));
19 vis[st]=1;
20 queue<int>que;
21 while(!que.empty()) que.pop();
22 que.push(st);
23 int u,v,flow=inf;
24 pre[st]=pre[ed]=-1;
25 while(!que.empty()){
26 u=que.front();
27 que.pop();
28 if( u==ed ) break;
29 for(int i=head[u];~i;i=e[i].nxt ){
30 v=e[i].to;
31 if(!vis[v]&&e[i].val>0){
32 vis[v]=1;
33 flow=min(flow,e[i].val);
34 pre[v]=i;//注意是i,不是u
35 que.push(v);
36 }
37 }
38 }
39 if( pre[ed]==-1 ) return -1;
40 return flow;
41 }
42 int EK(int st,int ed){
43 int stream,flow=0;
44 while((stream=BFS(st,ed))!=-1 ){
45 //pre通过节点存边的坐标,通过边的反向边的to获取上一个节点
46 for(int i=pre[ed];~i;i=pre[e[i^1].to]){
47 e[i].val-=stream;
48 e[i^1].val+=stream;
49 }
50 flow+=stream;
51 }
52 return flow;
53 }
54 int main()
55 {
56 int a,b,n,m,flow;
57 while(~scanf("%d%d",&m,&n)){
58 memset(head,-1,sizeof(head));
59 cnt=0;
60 while( m-- ){
61 scanf("%d%d%d",&a,&b,&flow);
62 addedge(a,b,flow);
63 addedge(b,a,0);
64 }
65 printf("%d\n",EK(1,n));
66 }
67 return 0;
68 }
View Code
FF+邻接矩阵：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int graph[maxn][maxn];
9 int vis[maxn];
10 int n,m;
11 int dfs(int st,int ed,int f){//在没有走到汇点前我们不知道流量是多少,所以f是动态更新的
12 vis[st]=1;
13 if(st==ed) return f;//走到汇点返回本次增广的流量
14 rep(i,1,n){
15 if( graph[st][i]>0 && vis[i]==0 ){
16 int d=dfs(i,ed,min(graph[st][i],f));
17 if( d>0 ){//顺着流过去,要受一路上最小容量的限制
18 graph[st][i]-=d;
19 graph[i][st]+=d;
20 return d;
21 }
22 }
23 }
24 return 0;//无法到汇点
25 }
26 int FF(int st,int ed){
27 int flow=0,curr;
28 while(1){
29 memset(vis,0,sizeof(vis));
30 curr=dfs(st,ed,inf);
31 if( curr==0 ) return flow;
32 flow+=curr;
33 }
34 }
35 int main()
36 {
37 int a,b,val;
38 while(~scanf("%d%d",&m,&n)){
39 memset(graph,0,sizeof(graph));
40 while(m--){
41 scanf("%d%d%d",&a,&b,&val);
42 graph[a][b]+=val;
43 }
44 printf("%d\n",FF(1,n));
45 }
46 return 0;
47 }
View Code
FF+邻接表：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 struct node{
9 int to;
10 int flow;
11 int rev;//反向边在数组中的位置
12 node(){}
13 node(int to,int flow,int rev):to(to),flow(flow),rev(rev){}
14 };
15 vector<node>graph[maxn];
16 int vis[maxn];
17 int n,m;
18 void addedge(int from,int to,int val){
19 graph[from].push_back(node(to,val,graph[to].size()));
20 graph[to].push_back(node(from,0,graph[from].size()-1));
21 }
22 int dfs(int st,int ed,int f){
23 vis[st]=1;
24 if(st==ed)return f;
25 rep(i,0,graph[st].size()-1){
26 node &temp=graph[st][i];//注意必须用引用
27 if(!vis[temp.to] && temp.flow>0 ){
28 int d=dfs(temp.to,ed,min(f,temp.flow));
29 if( d>0 ){
30 temp.flow-=d;
31 graph[temp.to][temp.rev].flow+=d;
32 return d;
33 }
34 }
35 }
36 return 0;
37 }
38 int FF(int st,int ed){
39 int flow=0,curr;
40 while(1){
41 memset(vis,0,sizeof(vis));
42 curr=dfs(st,ed,inf);
43 if( curr==0 ) return flow;
44 flow+=curr;
45 }
46 }
47 int main()
48 {
49 int a,b,val;
50 while(~scanf("%d%d",&m,&n)){
51 rep(i,1,n) graph[i].clear();
52 while( m-- ){
53 scanf("%d%d%d",&a,&b,&val);
54 addedge(a,b,val);
55 }
56 printf("%d\n",FF(1,n));
57 }
58 return 0;
59 }
View Code
FF+前向星：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 struct node{int to,nxt,c;}e[maxn<<1];
9 int head[maxn<<1],cnt,vis[maxn];
10 void addedge(int u,int v,int val){
11 e[cnt].to=v;
12 e[cnt].c=val;
13 e[cnt].nxt=head[u];
14 head[u]=cnt++;
15 }
16 int dfs(int st,int ed,int f){
17 vis[st]=1;
18 if(st==ed) return f;
19 for(int i=head[st];~i;i=e[i].nxt){
20 int v=e[i].to,val=e[i].c;
21 if( !vis[v]&&val>0 ){
22 int d=dfs(v,ed,min(val,f));
23 if( d>0 ){
24 e[i].c-=d;
25 e[i^1].c+=d;
26 return d;
27 }
28 }
29 }
30 return 0;
31 }
32 int FF(int st,int ed){
33 int flow=0,d;
34 while(1){
35 memset(vis,0,sizeof(vis));
36 d=dfs(st,ed,inf);
37 if(d==0) return flow;
38 flow+=d;
39 }
40 }
41 int main()
42 {
43 int n,m,a,b,val;
44 while(~scanf("%d%d",&m,&n)){
45 memset(head,-1,sizeof(head));
46 cnt=0;
47 while( m-- ){
48 scanf("%d%d%d",&a,&b,&val);
49 addedge(a,b,val);
50 addedge(b,a,0);
51 }
52 printf("%d\n",FF(1,n));
53 }
54 return 0;
55 }
View Code
Dinic+邻接矩阵：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int graph[maxn][maxn];
9 int step[maxn];//相当于将图分层了
10 int n,m;
11 bool bfs(int st,int ed){//bool 函数是一个小优化，判断是否能搜到汇点,如果连汇点都搜不到还dfs干什么
12 memset(step,-1,sizeof(step));
13 queue<int>que;
14 while(!que.empty()) que.pop();
15 step[st]=0;
16 que.push(st);
17 while(!que.empty()){
18 int curr=que.front();
19 que.pop();
20 rep(i,1,n){
21 if(graph[curr][i]>0 && step[i]==-1 ){
22 step[i]=step[curr]+1;
23 if( i==ed ) return true;
24 que.push(i);
25 }
26 }
27 }
28 return step[ed]!=-1;
29 }
30 int dfs(int st,int ed,int f){
31 if(st==ed||f==0) return f;
32 int flow=0,d;
33 rep(i,1,n){
34 if((step[i]==step[st]+1)&&graph[st][i]>0&&(d=dfs(i,ed,min(graph[st][i],f)))){//step[i]==step[st]+1;控制仅允许流向下一层
35 graph[st][i]-=d;
36 graph[i][st]+=d;
37 flow+=d;//累加当前节点的某条路径的合适流量
38 f-=d;//当前节点的容量减去某条路径的合适流量
39 if(f==0) break;//如果当前节点的容量用完,说明无法再通过任何流量
40 }
41 }
42 if( flow==0 ) step[st]=inf;//如果当前节点无任何流量通过,取消标记
43 return flow;
44 }
45 int Dinic(int st,int ed){
46 int flow=0;
47 while(bfs(st,ed)){
48 flow+=dfs(st,ed,inf);
49 }
50 return flow;
51 }
52 int main()
53 {
54 int a,b,flow;
55 while(~scanf("%d%d",&m,&n)){
56 memset(graph,0,sizeof(graph));
57 while(m--){
58 scanf("%d%d%d",&a,&b,&flow);
59 graph[a][b]+=flow;
60 }
61 printf("%d\n",Dinic(1,n));
62 }
63 return 0;
64 }
View Code
Dinic+邻接表：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int step[maxn];
9 struct node{
10 int to;
11 int flow;
12 int rev;//记录反向边的index
13 node(){}
14 node(int to,int flow,int rev):to(to),flow(flow),rev(rev){}
15 };
16 vector<node>graph[maxn];
17 int n,m;
18 void addedge(int from,int to,int flow){
19 graph[from].push_back(node(to,flow,graph[to].size()));
20 graph[to].push_back(node(from,0,graph[from].size()-1));
21 }
22 bool bfs(int st,int ed){
23 memset(step,-1,sizeof(step));
24 queue<int>que;
25 que.push(st);
26 step[st]=0;
27 while(!que.empty()){
28 int curr=que.front();
29 que.pop();
30 rep(i,0,graph[curr].size()-1){
31 node &temp=graph[curr][i];
32 if(step[temp.to]==-1 && temp.flow>0 ){
33 step[temp.to]=step[curr]+1;
34 if( temp.to==ed ) return true;
35 que.push(temp.to);
36 }
37 }
38 }
39 return step[ed]!=-1;
40 }
41 int dfs(int st,int ed,int f){
42 if( st==ed || f==0 ) return f;
43 int flow=0;
44 rep(i,0,graph[st].size()-1){
45 node &temp=graph[st][i];
46 if( temp.flow>0 && step[temp.to]==step[st]+1){
47 int d=dfs(temp.to,ed,min(f,temp.flow));
48 if(d>0){
49 temp.flow-=d;
50 graph[temp.to][temp.rev].flow+=d;
51 flow+=d;
52 f-=d;
53 if(f==0) break;
54 }
55 }
56 }
57 if( flow==0 ) step[st]=inf;
58 return flow;
59 }
60 int Dinic(int st,int ed){
61 int flow=0;
62 while(bfs(st,ed)){
63 flow+=dfs(st,ed,inf);
64 }
65 return flow;
66 }
67 int main()
68 {
69 int a,b,flow;
70 while(~scanf("%d%d",&m,&n)){
71 memset(graph,0,sizeof(graph));
72 while( m-- ){
73 scanf("%d%d%d",&a,&b,&flow);
74 addedge(a,b,flow);
75 }
76 printf("%d\n",Dinic(1,n));
77 }
78 return 0;
79 }
View Code
Dinic+前向星：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int head[maxn<<1],cnt,step[maxn];
9 struct node{int to,nxt,c;}e[maxn<<1];
10 void addedge(int u,int v,int val){
11 e[cnt].to=v;
12 e[cnt].c=val;
13 e[cnt].nxt=head[u];
14 head[u]=cnt++;
15 }
16 bool bfs(int st,int ed){
17 memset(step,-1,sizeof(step));
18 queue<int>que;
19 while(!que.empty())que.pop();
20 que.push(st);
21 step[st]=0;
22 int u;
23 while(!que.empty()){
24 u=que.front();
25 que.pop();
26 for(int i=head[u];~i;i=e[i].nxt){
27 int v=e[i].to;
28 if( step[v]==-1 && e[i].c>0 ){
29 step[v]=step[u]+1;
30 que.push(v);
31 if( v==ed ) return true;
32 }
33 }
34 }
35 return step[ed]!=-1;
36 }
37
38 int dfs(int st,int ed,int flow){
39 if(st==ed||flow==0) return flow;
40 int curr=0;
41 for(int i=head[st];~i;i=e[i].nxt){
42 int v=e[i].to;
43 int val=e[i].c;
44 if( step[st]+1==step[v] && val>0 ){
45 int d=dfs(v,ed,min(val,flow));
46 if(d>0){
47 e[i].c-=d;
48 e[i^1].c+=d;//添反向边的时候是相邻的,i^1取相邻数;
49 curr+=d;//累加当前节点的某条路径的合适流量
50 flow-=d;//当前节点的容量减去某条路径的合适流量
51 if(flow==0) break;//如果当前节点的容量用完,说明无法再通过任何流量
52 }
53 }
54 }
55 if( curr==0 ) step[st]=inf;//如果当前节点无任何流量通过,取消标记
56 return curr;
57 }
58
59 int Dinic(int st,int ed){
60 int flow=0;
61 while(bfs(st,ed)){
62 flow+=dfs(st,ed,inf);
63 }
64 return flow;
65 }
66 int main()
67 {
68 int n,m,a,b,val;
69 while(~scanf("%d%d",&m,&n)){
70 memset(head,-1,sizeof(head));
71 cnt=0;
72 while(m--){
73 scanf("%d%d%d",&a,&b,&val);
74 addedge(a,b,val);
75 addedge(b,a,0);
76 }
77 printf("%d\n",Dinic(1,n));
78 }
79 return 0;
80 }
View Code
Dinic+前向星+优化：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn=210;
5 const int inf=0x3f3f3f3f;
6 #define rep(i,first,last) for(int i=first;i<=last;i++)
7 #define dep(i,first,last) for(int i=first;i>=last;i--)
8 int head[maxn<<1],cnt,step[maxn],cur[maxn];
9 struct node{int to,nxt,c;}e[maxn<<1];
10 void addedge(int u,int v,int val){
11 e[cnt].to=v;
12 e[cnt].c=val;
13 e[cnt].nxt=head[u];
14 head[u]=cnt++;
15 }
16 bool bfs(int st,int ed){
17 memset(step,-1,sizeof(step));
18 queue<int>que;
19 while(!que.empty())que.pop();
20 que.push(st);
21 step[st]=0;
22 int u;
23 while(!que.empty()){
24 u=que.front();
25 que.pop();
26 for(int i=head[u];~i;i=e[i].nxt){
27 int v=e[i].to;
28 if( step[v]==-1 && e[i].c>0 ){
29 step[v]=step[u]+1;
30 que.push(v);
31 if( v==ed ) return true;
32 }
33 }
34 }
35 return step[ed]!=-1;
36 }
37
38 int dfs(int st,int ed,int flow){
39 if(st==ed||flow==0) return flow;
40 int curr=0;
41 for(int &i=cur[st];~i;i=e[i].nxt){
42 int v=e[i].to;
43 int val=e[i].c;
44 if( step[st]+1==step[v] && val>0 ){
45 int d=dfs(v,ed,min(val,flow));
46 if(d>0){
47 e[i].c-=d;
48 e[i^1].c+=d;
49 curr+=d;
50 flow-=d;
51 if(flow==0) break;
52 }
53 }
54 }
55 if( curr==0 ) step[st]=inf;
56 return curr;
57 }
58
59 int Dinic(int st,int ed){
60 int flow=0;
61 while(bfs(st,ed)){
62 rep(i,0,maxn-2) cur[i]=head[i];
63 flow+=dfs(st,ed,inf);
64 }
65 return flow;
66 }
67 int main()
68 {
69 int n,m,a,b,val;
70 while(~scanf("%d%d",&m,&n)){
71 memset(head,-1,sizeof(head));
72 cnt=0;
73 while(m--){
74 scanf("%d%d%d",&a,&b,&val);
75 addedge(a,b,val);
76 addedge(b,a,0);
77 }
78 printf("%d\n",Dinic(1,n));
79 }
80 return 0;
81 }
View Code
标签：流入,int,Ditches,flow,st,ed,graph,Drainage,return 来源： https://www.cnblogs.com/wsy107316/p/12447235.html