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题目链接:https://uva.onlinejudge.org/external/16/1658.pdf
题意:
n个地点m条边,有两个人要从1走到n,但是不能经过相同的地点(除了1和n),每条边都有一个权值,问两个人走到地点n花的总权值最小是多少。
做法:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=400000;
const int maxm=1000000;
const int inf=0x3f3f3f3f;
int dis[maxn];
int vis[maxn],pre[maxn];
int head[maxn],cnt,viss[maxn],visss[maxm];
int n,m,sp,tp,k,W;
ll ans=0;
struct node{
int to,cap,cost,next;
}e[maxm];
void add(int from,int to,int cap,int cost){
e[cnt].to=to; e[cnt].cap=cap;
e[cnt].cost=cost; e[cnt].next=head[from];
head[from]=cnt++;
e[cnt].to=from; e[cnt].cap=0;
e[cnt].cost=-cost; e[cnt].next=head[to];
head[to]=cnt++;
}
bool spfa(int s,int t,int &flow,int &cost){
queue<int> q;
memset(dis,inf,sizeof(dis));
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
dis[s]=0; q.push(s);
vis[s]=1;
int d=inf;
while(!q.empty()){
int u=q.front();
q.pop();
vis[u]=0;
for(int i=head[u];~i;i=e[i].next){
int v=e[i].to;
if(e[i].cap>0&&dis[v]>dis[u]+e[i].cost){
dis[v]=dis[u]+e[i].cost;
pre[v]=i;
if(!vis[v]){
vis[v]=1;
q.push(v);
}
}
}
}
if(dis[t]==inf){
return false;
}
for(int i=pre[t];~i;i=pre[e[i^1].to]){
d=min(d,e[i].cap);
}
for(int i=pre[t];~i;i=pre[e[i^1].to]){
e[i].cap-=d;
e[i^1].cap+=d;
cost+=e[i].cost*d;
}
flow+=d;
return true;
}
int mcmf(int s,int t){
int flow=0,cost=0;
while(spfa(s,t,flow,cost)){
//cout<<flow<<" "<<cost<<endl;
}
return cost;
}
int gainp(int m,int n){
return (m-1)*2+n;
}
int main(){
while(~scanf("%d%d",&n,&m)){
memset(head,-1,sizeof(head));
cnt=0;
sp=2*n+1;tp=2*n+2;
add(sp,gainp(1,0),2,0);
add(gainp(n,1),tp,2,0);
for(int i=0,x,y,z;i<m;i++){
scanf("%d%d%d",&x,&y,&z);
x=gainp(x,1);
y=gainp(y,0);
add(x,y,1,z);
}
for(int i=1;i<=n;i++){
int adds=1;
if(i==1||i==n) adds++;
add(gainp(i,0),gainp(i,1),adds,0);
}
cout<<mcmf(sp,tp)<<endl;
}
return 0;
}