http://poj.org/problem?id=1273
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions:87219 | Accepted: 33916 |
Description
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Input
Output
Sample Input
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10
Sample Output
50
现在有m个池塘(从1到m开始编号,1为源点,m为汇点),及n条水渠,给出这n条水渠所连接的池塘和所能流过的水量,求水渠中所能流过的水的最大容量.一道基础的最大流题目。
——其他练习题 POJ3436 、
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <bitset> 5 #include <cmath> 6 #include <cctype> 7 #include <iostream> 8 #include <algorithm> 9 #include <string> 10 #include <vector> 11 #include <queue> 12 #include <map> 13 #include <set> 14 #include <sstream> 15 #include <iomanip> 16 using namespace std; 17 typedef long long ll; 18 typedef unsigned long long ull; 19 const ll inff = 0x3f3f3f3f3f3f3f3f; 20 #define FOR(i,a,b) for(int i(a);i<=(b);++i) 21 #define FOL(i,a,b) for(int i(a);i>=(b);--i) 22 #define REW(a,b) memset(a,b,sizeof(a)) 23 #define inf int(0x3f3f3f3f) 24 #define si(a) scanf("%d",&a) 25 #define sl(a) scanf("%I64d",&a) 26 #define sd(a) scanf("%lf",&a) 27 #define ss(a) scanf("%s",a) 28 #define mod ll(998244353) 29 #define pb push_back 30 #define eps 1e-6 31 #define lc d<<1 32 #define rc d<<1|1 33 #define Pll pair<ll,ll> 34 #define P pair<int,int> 35 #define pi acos(-1) 36 const int N=100008,M=100008; 37 int head[N],tot,n,m,a,b,c; 38 struct node{ 39 int next,c,to;}e[M]; 40 struct max_flow{ 41 int S,T,n; 42 int lev[N],q[N],cur[N],f; 43 void init(int _s,int _t) 44 { 45 tot=0;S=_s,T=_t,n=T+1; 46 FOR(i,0,n) head[i]=-1; 47 } 48 void add(int a,int b,int c) 49 { 50 e[tot].next=head[a]; 51 e[tot].to=b; 52 e[tot].c=c; 53 head[a]=tot++; 54 } 55 void Add(int a,int b,int c) 56 { 57 add(a,b,c); 58 add(b,a,0); 59 } 60 int bfs() 61 { 62 FOR(i,0,n) lev[i]=0; 63 lev[S]=1,f=0,q[f++]=S; 64 FOR(i,0,f-1) 65 { 66 int u=q[i]; 67 for(int i=head[u];i!=-1;i=e[i].next) 68 if(lev[e[i].to]==0&&e[i].c>0) 69 { 70 int to=e[i].to; 71 lev[to]=lev[u]+1; 72 q[f++]=to; 73 if(to==T) return 1; 74 } 75 } 76 return 0; 77 } 78 int dfs(int u,int f) 79 { 80 if(u==T) return f; 81 int tag=0,c; 82 for(int &i=cur[u];i!=-1;i=e[i].next) 83 { 84 int to=e[i].to; 85 if(e[i].c>0&&lev[to]==lev[u]+1) 86 { 87 c=dfs(to,min(f-tag,e[i].c)); 88 e[i].c-=c; 89 e[i^1].c+=c; 90 tag+=c; 91 if(tag==f) return tag; 92 } 93 } 94 return tag; 95 } 96 int slove() 97 { 98 int ans=0; 99 while(bfs()) 100 { 101 FOR(i,0,n) cur[i]=head[i]; 102 ans+=dfs(S,inf); 103 } 104 return ans; 105 } 106 }flow; 107 int main() 108 { 109 cin.tie(0); 110 cout.tie(0); 111 while(~scanf("%d%d",&m,&n)) 112 { 113 flow.init(1,n); 114 while(m--) 115 { 116 si(a),si(b),si(c); 117 flow.Add(a,b,c); 118 } 119 cout<<flow.slove()<<endl; 120 } 121 return 0; 122 }
https://blog.csdn.net/huzhengnan/article/details/7766446
最大流的一些板子
https://blog.csdn.net/wjf_wzzc/article/details/24820525
sap:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdlib> 6 #include<string> 7 #include<queue> 8 #include<algorithm> 9 using namespace std; 10 11 const int N=100010; 12 const int M=400010; 13 const int inf=0xfffffff; 14 15 int n,m,cnt; 16 17 struct Edge{ 18 int v , cap , next; 19 } edge[M]; 20 21 int head[N],pre[N],d[N],numd[N];//分别为链表的头指针，每个点的前驱，每个点的d值，以及标号为d[i] 的点的个数 22 int cur_edge[N];//从每个点出发满足d[i] = d[j] + 1的边的地址 ， 插入边时的计数，源点与汇点 23 24 void addedge(int u,int v,int c){ 25 26 edge[cnt].v = v; 27 edge[cnt].cap = c; 28 edge[cnt].next = head[u]; 29 head[u] = cnt++; 30 31 edge[cnt].v = u; 32 edge[cnt].cap = 0; 33 edge[cnt].next = head[v]; 34 head[v] = cnt++; 35 } 36 37 void bfs(int s){ //先用广度优先算出每个点的d值 38 memset(numd,0,sizeof(numd)); 39 for(int i=1; i<=n; i++) 40 numd[ d[i] = n ]++; 41 d[s] = 0; 42 numd[n]--; 43 numd[0]++; 44 queue<int> Q; 45 Q.push(s); 46 47 while(!Q.empty()){ 48 int v=Q.front(); 49 Q.pop(); 50 51 int i=head[v]; 52 while(i != -1){ 53 int u=edge[i].v; 54 55 if(d[u]<n){ 56 i=edge[i].next; 57 continue ; 58 } 59 60 d[u]=d[v]+1; 61 numd[n]--; 62 numd[d[u]]++; 63 Q.push(u); 64 i=edge[i].next; 65 } 66 } 67 } 68 69 int SAP(int s,int t){ 70 for(int i = 1; i <= n; i++) 71 cur_edge[i] = head[i]; //当前满足d[i] = d[j] + 1的边的为第一条边 72 int max_flow=0; 73 bfs(t); 74 int u = s ;//从源点搜一条到汇点的增广路 75 while(d[s]<n){//就算所有的点连成一条线源点的d值也是最多是n-1 76 if(u == t){//如果找到一条增广路径 77 int cur_flow = inf,neck;//找到那条瓶颈边 78 for(int from = s; from != t; from = edge[cur_edge[from]].v){ 79 if(cur_flow > edge[cur_edge[from]].cap){ 80 neck = from; 81 cur_flow = edge[cur_edge[from]].cap; 82 } 83 } 84 85 for(int from = s; from != t; from = edge[cur_edge[from]].v){ //修改增广路上的边的容量 86 int tmp = cur_edge[from]; 87 edge[tmp].cap -= cur_flow; 88 edge[tmp^1].cap += cur_flow; 89 } 90 max_flow += cur_flow;//累加计算最大流 91 u = neck;//下一次搜索直接从瓶颈边的前一个节点搜起 92 } 93 94 int i; 95 for(i = cur_edge[u]; i != -1; i = edge[i].next) //从当前点开始找一条允许弧 96 if(edge[i].cap && d[u] == d[edge[i].v]+1)//如果找到跳出循环 97 break; 98 99 if(i!=-1){//找到一条允许弧 100 cur_edge[u] = i;//从点u出发的允许弧的地址 101 pre[edge[i].v] = u;//允许弧上下一个点的前驱为u 102 u = edge[i].v;//u变成下一个点继续搜直到搜出一条增广路 103 } 104 else{//如果没有搜到允许弧 105 numd[d[u]]--; //d[u]将被修改所以numd[d[u]]减一 106 if(!numd[d[u]]) break; //如果没有点的d值为d[u]则不可能再搜到增广路结束搜索 107 cur_edge[u] = head[u]; //当前点的允许弧为第一条边 108 int tmp = n; 109 for(int j = head[u]; j != -1; j = edge[j].next) //搜与u相连的点中d值最小的 110 if(edge[j].cap && tmp > d[edge[j].v]) 111 tmp = d[edge[j].v]; 112 113 d[u] = tmp+1; //修改d[u] 114 numd[d[u]]++; 115 if(u != s) 116 u = pre[u];//从u的前驱搜，因为从u没有搜到允许弧 117 } 118 } 119 return max_flow; 120 } 121 inline void pre_init(){ 122 cnt = 0; 123 memset(head, -1, sizeof head); 124 } 125 126 void mapping(){ 127 int u, v, w; 128 for(int i = 1; i <= m; ++i){ 129 scanf("%d %d %d", &u, &v, &w); 130 addedge(u, v, w); 131 } 132 } 133 134 int main(){ 135 while(~scanf("%d%d",&m,&n)){ 136 pre_init(); 137 mapping(); 138 int s,t; 139 s = 1;t = n; 140 int ans = SAP(s,t); 141 printf("%d\n",ans); 142 } 143 return 0; 144 }
dinic:
1 //dinic 2 3 #include <algorithm> 4 #include <iostream> 5 #include <string.h> 6 #include <cstdio> 7 #include <queue> 8 using namespace std; 9 const int inf = 0x3f3f3f3f; 10 const int maxn = 205; 11 const int maxm = maxn*maxn; 12 struct node{int w; int v, next;} edge[maxm]; 13 int pre[maxn], rec[maxn], head[maxn], block[maxn]; 14 int dis[maxn]; 15 int n, m, no; 16 int S, T; 17 queue<int> q; 18 inline void init(){ 19 no = 0; 20 memset(head, -1, sizeof head); 21 } 22 inline void add(int u, int v, int w){ 23 edge[no].v = v; edge[no].w = w; 24 edge[no].next = head[u]; head[u] = no++; 25 edge[no].v = u; edge[no].w = 0; 26 edge[no].next = head[v]; head[v] = no++; 27 } 28 void reset(int S, int T){ 29 memset(dis, 0x3f, sizeof dis); 30 memset(block, 0, sizeof block); 31 q.push(S); dis[S] = 0; 32 while(!q.empty()){ 33 int top = q.front(); q.pop(); 34 for(int k = head[top]; k != -1; k = edge[k].next) 35 if(dis[edge[k].v] == inf && edge[k].w) 36 dis[edge[k].v] = dis[top]+1, q.push(edge[k].v); 37 } 38 } 39 int dinic(int S, int T){ 40 int ans = 0, flow = inf; 41 int top = S; 42 reset(S, T); pre[S] = S; 43 while(dis[T] != inf){ 44 int k, tmp; 45 for(k = head[top]; k != -1; k = edge[k].next){ 46 if(edge[k].w && dis[edge[k].v]==dis[top]+1 && 47 !block[edge[k].v]) break; 48 } 49 if(k != -1){ 50 tmp = edge[k].v; 51 flow = min(flow, edge[k].w); 52 pre[tmp] = top, rec[tmp] = k; 53 top = tmp; 54 if(top == T){ 55 ans += flow; tmp = -1; 56 for(; top != S; top = pre[top]){ 57 edge[rec[top]].w -= flow; 58 edge[rec[top]^1].w += flow; 59 if(!edge[rec[top]].w) tmp = top; 60 } 61 flow = inf; 62 if(tmp != -1){ 63 top = pre[tmp]; 64 for(; top != S; top = pre[top]) 65 flow = min(flow, edge[rec[top]].w); 66 top = pre[tmp]; 67 } 68 } 69 } 70 else{ 71 block[top] = 1; 72 top = pre[top]; 73 if(block[S]) reset(S, T); 74 } 75 } 76 return ans; 77 } 78 void mapping(){ 79 int u, v, w; 80 for(int i = 1; i <= m; ++i){ 81 scanf("%d %d %d", &u, &v, &w); 82 add(u, v, w); 83 } 84 } 85 int main(){ 86 while(~scanf("%d %d", &m, &n)){ 87 S = 1, T = n; 88 init(); 89 mapping(); 90 printf("%d\n", dinic(S, T)); 91 } 92 return 0; 93 }
感谢https://www.cnblogs.com/Asumi/p/9751117.html
【最大流之Dinic算法】POJ1273 【 & 当前弧优化 & 】
https://www.cnblogs.com/DF-yimeng/p/8583698.html
//dinic
#include <algorithm>#include <iostream>#include <string.h>#include <cstdio>#include <queue>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 205;const int maxm = maxn*maxn;struct node{int w; int v, next;} edge[maxm];int pre[maxn], rec[maxn], head[maxn], block[maxn];int dis[maxn];int n, m, no;int S, T;queue<int> q;inline void init(){ no = 0; memset(head, -1, sizeof head);}inline void add(int u, int v, int w){ edge[no].v = v; edge[no].w = w; edge[no].next = head[u]; head[u] = no++; edge[no].v = u; edge[no].w = 0; edge[no].next = head[v]; head[v] = no++;}void reset(int S, int T){ memset(dis, 0x3f, sizeof dis); memset(block, 0, sizeof block); q.push(S); dis[S] = 0; while(!q.empty()){ int top = q.front(); q.pop(); for(int k = head[top]; k != -1; k = edge[k].next) if(dis[edge[k].v] == inf && edge[k].w) dis[edge[k].v] = dis[top]+1, q.push(edge[k].v); }}int dinic(int S, int T){ int ans = 0, flow = inf; int top = S; reset(S, T); pre[S] = S; while(dis[T] != inf){ int k, tmp; for(k = head[top]; k != -1; k = edge[k].next){ if(edge[k].w && dis[edge[k].v]==dis[top]+1 && !block[edge[k].v]) break; } if(k != -1){ tmp = edge[k].v; flow = min(flow, edge[k].w); pre[tmp] = top, rec[tmp] = k; top = tmp; if(top == T){ ans += flow; tmp = -1; for(; top != S; top = pre[top]){ edge[rec[top]].w -= flow; edge[rec[top]^1].w += flow; if(!edge[rec[top]].w) tmp = top; } flow = inf; if(tmp != -1){ top = pre[tmp]; for(; top != S; top = pre[top]) flow = min(flow, edge[rec[top]].w); top = pre[tmp]; } } } else{ block[top] = 1; top = pre[top]; if(block[S]) reset(S, T); } } return ans;}void mapping(){ int u, v, w; for(int i = 1; i <= m; ++i){ scanf("%d %d %d", &u, &v, &w); add(u, v, w); }}int main(){ while(~scanf("%d %d", &m, &n)){ S = 1, T = n; init(); mapping(); printf("%d\n", dinic(S, T)); } return 0;}