hdu 1853 Cyclic Tour (二分匹配KM最小权值 或 最小费用最大流)
Cyclic Tour
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1197 Accepted Submission(s): 626
Problem Description
There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?
Input
There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).
Output
Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.
Sample Input
6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1
Sample Output
42
-1
Hint
In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.
Author
RoBa@TJU
Source
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题意：
给一个图，要求用若干个环遍历全部点，求最短路程。
二分匹配KM最小权值做法：
1 //46MS 284K 2108 B C++ 2 /* 3 二分匹配： 4 KM算法最小权值模板题... 5 */ 6 #include<stdio.h> 7 #include<string.h> 8 #define inf 0x7ffffff 9 int g[105][105]; 10 int slack[105]; 11 int match[105]; 12 int lx[105],ly[105],visx[105],visy[105]; 13 int n,m; 14 int Min(int a,int b) 15 { 16 return a<b?a:b; 17 } 18 int Max(int a,int b) 19 { 20 return a>b?a:b; 21 } 22 int dfs(int x) 23 { 24 visx[x]=1; 25 for(int i=1;i<=n;i++){ 26 if(!visy[i]){ 27 if(lx[x]+ly[i]==g[x][i]){ 28 visy[i]=1; 29 if(match[i]==-1 || dfs(match[i])){ 30 match[i]=x; 31 return 1; 32 } 33 }else{ 34 slack[i]=Min(slack[i],lx[x]+ly[i]-g[x][i]); 35 } 36 } 37 } 38 return 0; 39 } 40 void KM() 41 { 42 memset(match,-1,sizeof(match)); 43 for(int i=0;i<=n;i++) lx[i]=-inf; 44 memset(ly,0,sizeof(ly)); 45 for(int i=1;i<=n;i++) 46 for(int j=1;j<=n;j++) 47 lx[i]=Max(lx[i],g[i][j]); 48 for(int i=1;i<=n;i++){ 49 for(int j=1;j<=n;j++) 50 slack[j]=inf; 51 while(true){ 52 memset(visx,0,sizeof(visx)); 53 memset(visy,0,sizeof(visy)); 54 if(dfs(i)) break; 55 int temp=inf; 56 for(int j=1;j<=n;j++) 57 if(!visy[j]) 58 temp=Min(temp,slack[j]); 59 for(int j=1;j<=n;j++){ 60 if(visx[j]) lx[j]-=temp; 61 if(visy[j]) ly[j]+=temp; 62 else slack[j]-=temp; 63 } 64 } 65 66 } 67 } 68 int main(void) 69 { 70 int a,b,c; 71 while(scanf("%d%d",&n,&m)!=EOF) 72 { 73 for(int i=0;i<=n;i++) 74 for(int j=0;j<=n;j++) 75 g[i][j]=-inf; 76 for(int i=0;i<m;i++){ 77 scanf("%d%d%d",&a,&b,&c); 78 if(-c>g[a][b]) 79 g[a][b]=-c; 80 } 81 KM(); 82 int ans=0; 83 for(int i=1;i<=n;i++){ 84 if(match[i]==-1 || g[match[i]][i]==-inf){ 85 ans=inf; 86 break; 87 } 88 ans+=g[match[i]][i]; 89 } 90 if(ans==inf) puts("-1"); 91 else printf("%d\n",-ans); 92 } 93 return 0; 94 } 95
最小费用最大流做法：
1 //218MS 656K 2377 B C++ 2 /* 3 4 第一题最小费用最大流...基本都是抄= = 5 注意边的初始化、添加和修改.. 6 7 */ 8 #include<iostream> 9 #include<queue> 10 #define inf 0x7ffffff 11 #define N 105 12 using namespace std; 13 struct node{ 14 int u,v,c,w; 15 int next; 16 }edge[4*N*N]; 17 int n,m,edgenum,sumflow; 18 int head[4*N],d[4*N],pp[4*N]; //pp记录增广链 19 bool vis[4*N]; 20 void init() //初始化 21 { 22 edgenum=0; 23 memset(head,-1,sizeof(head)); 24 } 25 void addedge(int u,int v,int c,int w) //添加双向边 26 { 27 edge[edgenum].u=u; 28 edge[edgenum].v=v; 29 edge[edgenum].c=c; 30 edge[edgenum].w=w; 31 edge[edgenum].next=head[u]; 32 head[u]=edgenum++; 33 edge[edgenum].u=v; 34 edge[edgenum].v=u; 35 edge[edgenum].c=0; //逆向边没流量 36 edge[edgenum].w=-w; //值取负 37 edge[edgenum].next=head[v]; 38 head[v]=edgenum++; 39 } 40 bool spfa() //求最短路 41 { 42 queue<int>Q; 43 memset(vis,false,sizeof(vis)); 44 memset(pp,-1,sizeof(pp)); 45 for(int i=0;i<=2*(n+1);i++) d[i]=inf; 46 vis[0]=true; 47 d[0]=0; 48 Q.push(0); 49 while(!Q.empty()){ 50 int u=Q.front(); 51 Q.pop(); 52 vis[u]=false; 53 for(int i=head[u];i!=-1;i=edge[i].next){ 54 int v=edge[i].v; 55 if(edge[i].c && d[v]>d[u]+edge[i].w){ 56 d[v]=d[u]+edge[i].w; 57 pp[v]=i; 58 if(!vis[v]){ 59 Q.push(v); 60 vis[v]=true; 61 } 62 } 63 } 64 } 65 if(d[2*n+1]==inf) return false; 66 return true; 67 } 68 int MCMF() 69 { 70 int t=2*n+1; 71 int flow=0; 72 int mincost=0; 73 sumflow=0; 74 while(spfa()){ 75 int minflow=inf+1; 76 for(int i=pp[t];i!=-1;i=pp[edge[i].u]) 77 if(edge[i].c<minflow) 78 minflow=edge[i].c; 79 flow+=minflow; 80 for(int i=pp[t];i!=-1;i=pp[edge[i].u]){ //调整增广链 81 edge[i].c-=minflow; 82 edge[i^1].c+=minflow; //逆向边 83 } 84 mincost+=d[t]*minflow; 85 } 86 sumflow=flow; 87 return mincost; 88 } 89 int main(void) 90 { 91 int a,b,c; 92 while(scanf("%d%d",&n,&m)!=EOF) 93 { 94 init(); 95 for(int i=1;i<=n;i++){ 96 addedge(0,i,1,0); 97 addedge(n+i,2*n+1,1,0); 98 } 99 for(int i=0;i<m;i++){ 100 scanf("%d%d%d",&a,&b,&c); 101 addedge(a,b+n,1,c); 102 } 103 int ans=MCMF(); 104 if(sumflow!=n) puts("-1"); 105 else printf("%d\n",ans); 106 } 107 return 0; 108 }
