hdu 1853(拆点判环+费用流)
Cyclic Tour
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 2257 Accepted Submission(s): 1148
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?
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).
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
-1
In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = ;
const int N = ;
const int M = ;
struct Edge{
int u,v,cap,cost,next;
}edge[M];
int head[N],tot,low[N],pre[N];
int total ;
bool vis[N];
void addEdge(int u,int v,int cap,int cost,int &k){
edge[k].u=u,edge[k].v=v,edge[k].cap = cap,edge[k].cost = cost,edge[k].next = head[u],head[u] = k++;
edge[k].u=v,edge[k].v=u,edge[k].cap = ,edge[k].cost = -cost,edge[k].next = head[v],head[v] = k++;
}
void init(){
memset(head,-,sizeof(head));
tot = ;
}
bool spfa(int s,int t,int n){
memset(vis,false,sizeof(vis));
for(int i=;i<=n;i++){
low[i] = (i==s)?:INF;
pre[i] = -;
}
queue<int> q;
q.push(s);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
for(int k=head[u];k!=-;k=edge[k].next){
int v = edge[k].v;
if(edge[k].cap>&&low[v]>low[u]+edge[k].cost){
low[v] = low[u] + edge[k].cost;
pre[v] = k; ///v为终点对应的边
if(!vis[v]){
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t]==-) return false;
return true;
}
int MCMF(int s,int t,int n){
int mincost = ,minflow,flow=;
while(spfa(s,t,n))
{
minflow=INF+;
for(int i=pre[t];i!=-;i=pre[edge[i].u])
minflow=min(minflow,edge[i].cap);
flow+=minflow;
for(int i=pre[t];i!=-;i=pre[edge[i].u])
{
edge[i].cap-=minflow;
edge[i^].cap+=minflow;
}
mincost+=low[t]*minflow;
}
total=flow;
return mincost;
}
int n,m;
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
init();
int src = ,des = *n+;
for(int i=;i<=n;i++){
addEdge(src,i,,,tot);
addEdge(i+n,des,,,tot);
}
for(int i=;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addEdge(u,v+n,,w,tot);
}
int mincost = MCMF(src,des,*n+);
if(total!=n) printf("-1\n");
else printf("%d\n",mincost);
}
}
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