题意: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.
求1到n的最大流，套模板。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 using namespace std; 7 #define inf 0x7fffffff 8 #define N 205 9 #define M 405 10 int cur[N],pre[N],gap[N],dis[N]; 11 int size,head[N]; 12 struct Edge 13 { 14 int v,w,next; 15 Edge(){} 16 Edge(int V,int W,int NEXT):v(V),w(W),next(NEXT){} 17 }edge[M]; 18 void Init() 19 { 20 memset(head,-1,sizeof(head)); 21 size = 0; 22 } 23 void InsertEdge(int u,int v,int w) 24 { 25 edge[size] = Edge(v,w,head[u]); 26 head[u] = size++; 27 edge[size] = Edge(u,0,head[v]); 28 head[v] = size++; 29 } 30 int Isap(int st,int ed,int n) 31 { 32 for(int i=0; i<=n; i++) 33 { 34 dis[i] = gap[i] = 0; 35 cur[i] = head[i]; 36 } 37 int u = pre[st] = st; 38 int aug = inf , maxflow = 0; 39 while(dis[st] < n) 40 { 41 loop: 42 for(int &i=cur[u]; i!=-1; i=edge[i].next) 43 { 44 int v = edge[i].v; 45 if(edge[i].w && dis[u] == dis[v] + 1) 46 { 47 aug = min(aug,edge[i].w); 48 pre[v] = u; 49 u = v; 50 if(v==ed) 51 { 52 maxflow += aug; 53 for(u=pre[u]; v!=st; v=u,u=pre[u]) 54 { 55 edge[cur[u]].w -= aug; 56 edge[cur[u]^1].w += aug; 57 } 58 aug = inf; 59 } 60 goto loop; 61 } 62 } 63 int mindis = n; 64 for(int i=head[u]; i!=-1; i=edge[i].next) 65 { 66 int v =edge[i].v; 67 if(edge[i].w && dis[v]<mindis) 68 { 69 cur[u] = i; 70 mindis = dis[v]; 71 } 72 } 73 if(--gap[dis[u]]==0) break; 74 gap[dis[u]=mindis +1]++; 75 u = pre[u]; 76 } 77 return maxflow; 78 } 79 80 int main() 81 { 82 int m,n,u,v,w,st,ed,nv; 83 while(scanf("%d%d",&m,&n)!=EOF) 84 { 85 Init(); 86 st = 1 ; ed = n ; nv = n; 87 for(int i=1; i<=m; i++) 88 { 89 scanf("%d%d%d",&u,&v,&w); 90 InsertEdge(u,v,w); 91 } 92 printf("%d ",Isap(st,ed,nv)); 93 } 94 return 0; 95 }