2023.1.12 有源汇上下界最大流简笔
警示
2023.1.12
[模板]有源汇上下界最大流
错误点：
1.建边方法简便：靠近于用u,v,w建边，不要冗杂多余（9-17行）
2.算法注意点：首先抽出每条边的下界(每条边边权 = 上界 - 下界)，建立虚拟源汇点进行dinic,答案得出后虚拟源点到汇点的流则为可行流，若虚拟源点的流出边有一个没有满流，则无解
其次在原图上进行dinic直到没有增广路为止，二答案相加则为最大流答案。
1 #include<bits/stdc++.h> 2 #define int long long 3 using namespace std; 4 const int N = 1e5 + 5,inf = 0x7fffffff; 5 struct Edge{ 6 int l,r,w,v,u,next; 7 }e[N * 2]; 8 int head[N],n,m,g[N],c,p,ans = 0,tot = 1,d[N],vis[N],nw[N],st = 1,ed = n + m + 2,in[N],out[N]; 9 inline void add(int x,int y,int w) 10 { 11 ++tot; 12 e[tot].u = x; 13 e[tot].w = w; 14 e[tot].v = y; 15 e[tot].next = head[x]; 16 head[x] = tot; 17 } 18 inline void init() 19 { 20 for(int i=1;i<=tot;i++) 21 { 22 if(e[i].l == 0 && e[i].r == 0) continue; 23 e[i].w = e[i].r - e[i].l; 24 } 25 for(int i=1;i<=n + m + 2;i++) 26 { 27 if(in[i] > out[i]) 28 { 29 add(n + m + 3,i,in[i] - out[i]); 30 e[tot].w = in[i] - out[i]; 31 add(i,n + m + 3,0); 32 } 33 if(out[i] > in[i]) 34 { 35 add(i,n + m + 4,out[i] - in[i]); 36 e[tot].w = out[i] - in[i]; 37 add(n + m + 4,i,0); 38 } 39 } 40 } 41 inline bool bfs() 42 { 43 for(int i=1;i<=n + m + 4;i++) d[i] = 0x7fffffff; 44 queue <int> q; 45 while(!q.empty()) q.pop(); 46 q.push(st); 47 d[st] = 1; 48 nw[st] = head[st]; 49 while(!q.empty()) 50 { 51 int now = q.front(); 52 q.pop(); 53 for(int i = head[now];i;i = e[i].next) 54 { 55 int to = e[i].v; 56 if(e[i].w <= 0 || d[to] < 0x7fffffff) continue; 57 nw[to] = head[to]; 58 d[to] = d[now] + 1; 59 q.push(to); 60 } 61 } 62 if(d[ed] < 0x7fffffff) return 1; 63 return 0; 64 } 65 inline int dinic(int x,int flow) 66 { 67 if(x == ed) return flow; 68 int rest = flow; 69 for(int i = nw[x];i && rest;i = e[i].next) 70 { 71 nw[x] = i; 72 int to = e[i].v; 73 if(d[to] != d[x] + 1 || e[i].w <= 0) continue; 74 int k = dinic(to,min(rest,e[i].w)); 75 if(!k) d[to] = inf; 76 rest -= k; 77 e[i].w -= k; 78 e[i ^ 1].w += k; 79 } 80 return flow - rest; 81 } 82 signed main() 83 { 84 while(cin>>n>>m) 85 { 86 memset(in,0,sizeof(in)); 87 memset(out,0,sizeof(out)); 88 tot = 1; 89 memset(head,0,sizeof(head)); 90 memset(nw,0,sizeof(nw)); 91 ans = 0; 92 int t,l,r; 93 for(int i=1;i<=m;i++) 94 { 95 cin>>g[i]; 96 } 97 for(int i=1;i<=n;i++) 98 { 99 cin>>c>>p; 100 for(int j=1;j<=c;j++) 101 { 102 cin>>t>>l>>r; 103 t++; 104 add(i + 1,n + t + 1,r - l); 105 add(n + t + 1,i + 1,0); 106 out[i + 1] += l; 107 in[n + t + 1] += l; 108 } 109 add(1,i + 1,p); 110 add(i + 1,1,0); 111 } 112 for(int i=1;i<=m;i++) 113 { 114 add(i + n + 1,n + m + 2,inf - g[i]); 115 add(n + m + 2,i + n + 1,0); 116 out[i + n + 1] += g[i]; 117 in[n + m + 2] += g[i]; 118 } 119 st = n + m + 3;ed = n + m + 4; 120 init(); 121 add(n + m + 2,1,inf); 122 add(1,n + m + 2,0); 123 int flow = 0; 124 while(bfs()) 125 while(flow = dinic(st,inf)) 126 flow = 0; 127 int flag = 1; 128 ans += (inf - e[tot - 1].w); 129 for(int i = head[st];i;i = e[i].next) 130 if(e[i].w > 0) 131 flag = 0; 132 if(flag == 0) 133 { 134 cout<<-1<<endl<<endl; 135 continue; 136 } 137 e[tot].w = 0; 138 e[tot - 1].w = 0; 139 st = 1; 140 ed = n + m + 2; 141 flow = 0; 142 while(bfs()) 143 while(flow = dinic(st,inf)) 144 ans += flow; 145 cout<<ans<<endl<<endl; 146 } 147 return 0; 148 }
