J: Our Journey of Dalian Ends

urumqi-2017-online

存一个学姐用的模板(:з」∠)

因为每个点只能经过一次，拆点，限制为1

以上海为起点

大连和西安两个点再连向终点，限制为1

判断最终流量是否为2就能判断是否从大连开始到西安结束了

#include<bits/stdc++.h>
using namespace std;
#define N 30005
#define M 60005
#define INF 10000000
#define ms(x,y) memset(x,y,sizeof(x))
const double EPS=1e-9;
struct Edge{
int u,v,cap,next;
double cost;
}edge[M<<1];
int sumFlow,edgeNum,head
,pp
;
double dist
;
bool inq
;
void add(int u,int v,int cap,double cost){
edge[edgeNum].u=u;edge[edgeNum].v=v;edge[edgeNum].cap=cap;edge[edgeNum].cost=cost;
edge[edgeNum].next=head[u];head[u]=edgeNum++;
edge[edgeNum].u=v;edge[edgeNum].v=u;edge[edgeNum].cap=0;edge[edgeNum].cost=-cost;
edge[edgeNum].next=head[v];head[v]=edgeNum++;
}
bool SPFA(int s,int t,int n){
int i,u,v;
queue<int>q;
ms(inq,0),ms(pp,-1);
for(i=0;i<=n;i++) dist[i]=INF;
inq[s]=1;dist[s]=0;
q.push(s);
while(!q.empty()){
u=q.front();q.pop();inq[u]=0;
for(i=head[u];i!=-1;i=edge[i].next){
v=edge[i].v;
if(edge[i].cap&&dist[v]>dist[u]+edge[i].cost+EPS){
dist[v]=dist[u]+edge[i].cost;
pp[v]=i;
if(!inq[v]){
q.push(v);
inq[v]=1;
}
}
}
}
if(pp[t]==-1) return false;
return true;
}
pair<int,double> MCMF(int s,int t,int n){
int flow=0,i,minflow;
double mincost=0;
while(SPFA(s,t,n)){
minflow=INF;
for(i=pp[t];i!=-1;i=pp[edge[i].u])
minflow=min(edge[i].cap,minflow);
flow+=minflow;
for(i=pp[t];i!=-1;i=pp[edge[i].u])
edge[i].cap-=minflow,edge[i^1].cap+=minflow;
mincost+=dist[t]*minflow;
}
return make_pair(flow,mincost);
}

map<string,int>MP;
int main()
{
int t,n,m,i,u
,v
,w
;
scanf("%d",&t);
while(t--){
edgeNum=0,ms(head,-1);

MP.clear();
MP["Shanghai"]=1;
MP["Dalian"]=2;
MP["Xian"]=n=3;

scanf("%d",&m);
string s;
for(i=1;i<=m;i++){
cin>>s;
if(MP.find(s)==MP.end()) MP[s]=++n;
u[i]=MP[s];
cin>>s;
if(MP.find(s)==MP.end()) MP[s]=++n;
v[i]=MP[s];
scanf("%d",&w[i]);
}
for(i=1;i<=m;i++)
if(u[i]==1) add(u[i],v[i],1,w[i]);
else if (v[i]==1) add(v[i],u[i],1,w[i]);
else add(u[i]+n-1,v[i],1,w[i]),add(v[i]+n-1,u[i],1,w[i]);

// add(0,1,2,0);
for(i=2;i<=n;i++) add(i,i+n-1,1,0);
add(2+n-1,2*n,1,0);
add(3+n-1,2*n,1,0);

pair<int,double> p=MCMF(1,2*n,2*n);
if(p.first==2) printf("%.0f\n",p.second);
else puts("-1");
}
}