标签:style blog http color os io
很容易发现是网络流的题目,但最少边怎么求呢?初时想不到,但画图后忽然发现可以这样:
求一次网络流最小割后,把满流的边置1,不满流的置INF。再求一次最大流即可。
为什么呢?
是否会存在一些边当前不满流,但有可能是最少边数最少割的边呢?否。因为按照DINIC的求法,每次都是增广容量最少的路,若当前不满流,则必定不是最小割的边,所以只需在满流的边,即可组成最小割的边寻找最少边就可以了。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 7 const int INF=0x3f3f3f3f; 8 const int MAXN=1050; 9 const int MAXM=400000; 10 11 struct Node{ 12 int from,to,next; 13 int cap; 14 }edge[MAXM]; 15 int tol; 16 17 int dep[MAXN]; 18 int head[MAXN]; 19 20 int n,m; 21 void init(){ 22 tol=0; 23 memset(head,-1,sizeof(head)); 24 } 25 void addedge(int u,int v,int w){ 26 edge[tol].from=u; 27 edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; 28 head[u]=tol++; 29 edge[tol].from=v; 30 edge[tol].to=u; 31 edge[tol].cap=0; 32 edge[tol].next=head[v]; 33 head[v]=tol++; 34 } 35 36 int BFS(int start,int end){ 37 int que[MAXN]; 38 int front,rear; front=rear=0; 39 memset(dep,-1,sizeof(dep)); 40 que[rear++]=start; 41 dep[start]=0; 42 while(front!=rear) { 43 int u=que[front++]; 44 if(front==MAXN)front=0; 45 for(int i= head[u];i!=-1; i=edge[i].next){ 46 int v=edge[i].to; 47 if(edge[i].cap>0&& dep[v]==-1){ 48 dep[v]=dep[u]+1; 49 que[rear++]=v; 50 if(rear>=MAXN) rear=0; 51 if(v==end)return 1; 52 } 53 } 54 } 55 return 0; 56 } 57 int dinic(int start,int end){ 58 int res=0; 59 int top; 60 int stack[MAXN]; 61 int cur[MAXN]; 62 while(BFS(start,end)){ 63 memcpy(cur,head, sizeof(head)); 64 int u=start; 65 top=0; 66 while(1){ 67 if(u==end){ 68 int min=INF; 69 int loc; 70 for(int i=0;i<top;i++) 71 if(min>edge [stack[i]].cap) { 72 min=edge [stack[i]].cap; 73 loc=i; 74 } 75 for(int i=0;i<top;i++){ 76 edge[stack[i]].cap-=min; 77 edge[stack[i]^1].cap+=min; 78 } 79 res+=min; 80 top=loc; 81 u=edge[stack[top]].from; 82 } 83 for(int i=cur[u]; i!=-1; cur[u]=i=edge[i].next) 84 if(edge[i].cap!=0 && dep[u]+1==dep[edge[i].to]) 85 break; 86 if(cur[u] !=-1){ 87 stack [top++]= cur[u]; 88 u=edge[cur[u]].to; 89 } 90 else{ 91 if(top==0) break; 92 dep[u]=-1; 93 u= edge[stack [--top] ].from; 94 } 95 } 96 } 97 return res; 98 } 99 100 int main(){ 101 int T,cas=0; 102 int u,v,w,d; 103 scanf("%d",&T); 104 while(T--){ 105 init(); 106 cas++; 107 scanf("%d%d",&n,&m); 108 for(int i=0;i<m;i++){ 109 scanf("%d%d%d%d",&u,&v,&w,&d); 110 if(d){ 111 addedge(v,u,w); 112 } 113 addedge(u,v,w); 114 } 115 dinic(0,n-1); 116 for(int i=0;i<tol;i+=2){ 117 if(edge[i].cap==0){ 118 edge[i].cap=1; edge[i^1].cap=0; 119 } 120 else { 121 edge[i].cap=INF; 122 edge[i^1].cap=0; 123 } 124 } 125 int ans=dinic(0,n-1); 126 printf("Case %d: ",cas); 127 printf("%d\n",ans); 128 } 129 return 0; 130 }
DINIC的模板
1 const int INF=0x3f3f3f3f; 2 const int MAXN=1050; 3 const int MAXM=400000; 4 5 struct Node{ 6 int from,to,next; 7 int cap; 8 }edge[MAXM]; 9 int tol; 10 11 int dep[MAXN]; 12 int head[MAXN]; 13 14 int n,m; 15 void init(){ 16 tol=0; 17 memset(head,-1,sizeof(head)); 18 } 19 void addedge(int u,int v,int w){ 20 edge[tol].from=u; 21 edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; 22 head[u]=tol++; 23 edge[tol].from=v; 24 edge[tol].to=u; 25 edge[tol].cap=0; 26 edge[tol].next=head[v]; 27 head[v]=tol++; 28 } 29 30 int BFS(int start,int end){ 31 int que[MAXN]; 32 int front,rear; front=rear=0; 33 memset(dep,-1,sizeof(dep)); 34 que[rear++]=start; 35 dep[start]=0; 36 while(front!=rear) { 37 int u=que[front++]; 38 if(front==MAXN)front=0; 39 for(int i= head[u];i!=-1; i=edge[i].next){ 40 int v=edge[i].to; 41 if(edge[i].cap>0&& dep[v]==-1){ 42 dep[v]=dep[u]+1; 43 que[rear++]=v; 44 if(rear>=MAXN) rear=0; 45 if(v==end)return 1; 46 } 47 } 48 } 49 return 0; 50 } 51 int dinic(int start,int end){ 52 int res=0; 53 int top; 54 int stack[MAXN]; 55 int cur[MAXN]; 56 while(BFS(start,end)){ 57 memcpy(cur,head, sizeof(head)); 58 int u=start; 59 top=0; 60 while(1){ 61 if(u==end){ 62 int min=INF; 63 int loc; 64 for(int i=0;i<top;i++) 65 if(min>edge [stack[i]].cap) { 66 min=edge [stack[i]].cap; 67 loc=i; 68 } 69 for(int i=0;i<top;i++){ 70 edge[stack[i]].cap-=min; 71 edge[stack[i]^1].cap+=min; 72 } 73 res+=min; 74 top=loc; 75 u=edge[stack[top]].from; 76 } 77 for(int i=cur[u]; i!=-1; cur[u]=i=edge[i].next) 78 if(edge[i].cap!=0 && dep[u]+1==dep[edge[i].to]) 79 break; 80 if(cur[u] !=-1){ 81 stack [top++]= cur[u]; 82 u=edge[cur[u]].to; 83 } 84 else{ 85 if(top==0) break; 86 dep[u]=-1; 87 u= edge[stack [--top] ].from; 88 } 89 } 90 } 91 return res; 92 }
HDU 3987 && DINIC,布布扣,bubuko.com
标签:style blog http color os io
原文地址:http://www.cnblogs.com/jie-dcai/p/3860050.html