描述
Alice and Bob are walking in an ancient maze with a lot of caves and one-way passages connecting them. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are too small for two people to walk through simultaneously, and crossing a passage can make it even more difficult to pass for the next person. We define di as the difficulty of crossing passage i for the first time, and ai as the additional difficulty for the second time (e.g. the second person's difficulty is di+ai).
Your task is to find two (possibly identical) routes for Alice and Bob, so that their total difficulty is minimized.
For
example, in figure 1, the best solution is 1->2->4 for both Alice
and Bob, but in figure 2, it's better to use 1->2->4 for Alice
and 1->3->4 for Bob.
输入
There
will be at most 200 test cases. Each case begins with two integers n, m
(1<=n<=500, 1<=m<=2000), the number of caves and passages.
Each of the following m lines contains four integers u, v, di and ai
(1<=u,v<=n, 1<=di<=1000, 0<=ai<=1000). Note that there
can be multiple passages connecting the same pair of caves, and even
passages connecting a cave and itself.
输出
For each test case, print the case number and the minimal total difficulty.
样例输入
4 4
1 2 5 1
2 4 6 0
1 3 4 0
3 4 9 1
4 4
1 2 5 10
2 4 6 10
1 3 4 10
3 4 9 10
样例输出
Case 1: 23
Case 2: 24
题意
找两条从1到N的路使得总花费最小，一条路第一次走花费d，第二次走花费d+a
题解
每个点建两条边一条流量1花费d，一条流量1花费d+a
源点S=0，和1连一条边流量2花费0
汇点T=n+1，和n连一条边流量2花费0
然后跑一边最小费用最大流
代码
#include<bits/stdc++.h>
using namespace std; const int N=1e5+;
const int M=2e5+;
const int INF=0x3f3f3f3f; int FIR[N],FROM[M],TO[M],CAP[M],FLOW[M],COST[M],NEXT[M],tote;
int pre[N],dist[N],q[];
bool vis[N];
int n,m,S,T;
void init()
{
tote=;
memset(FIR,-,sizeof(FIR));
}
void addEdge(int u,int v,int cap,int cost)
{
FROM[tote]=u;
TO[tote]=v;
CAP[tote]=cap;
FLOW[tote]=;
COST[tote]=cost;
NEXT[tote]=FIR[u];
FIR[u]=tote++; FROM[tote]=v;
TO[tote]=u;
CAP[tote]=;
FLOW[tote]=;
COST[tote]=-cost;
NEXT[tote]=FIR[v];
FIR[v]=tote++;
}
bool SPFA(int s, int t)
{
memset(dist,INF,sizeof(dist));
memset(vis,false,sizeof(vis));
memset(pre,-,sizeof(pre));
dist[s] = ;vis[s]=true;q[]=s;
int head=,tail=;
while(head!=tail)
{
int u=q[++head];vis[u]=false;
for(int v=FIR[u];v!=-;v=NEXT[v])
{
if(dist[TO[v]]>dist[u]+COST[v]&&CAP[v]>FLOW[v])
{
dist[TO[v]]=dist[u]+COST[v];
pre[TO[v]]=v;
if(!vis[TO[v]])
{
vis[TO[v]] = true;
q[++tail]=TO[v];
}
}
}
}
return pre[t]!=-;
}
void MCMF(int s, int t, int &cost, int &flow)
{
flow=;
cost=;
while(SPFA(s,t))
{
int Min = INF;
for(int v=pre[t];v!=-;v=pre[TO[v^]])
Min = min(Min, CAP[v]-FLOW[v]);
for(int v=pre[t];v!=-;v=pre[TO[v^]])
{
FLOW[v]+=Min;
FLOW[v^]-=Min;
cost+=COST[v]*Min;
}
flow+=Min;
}
}
int main()
{
int ca=;
while(scanf("%d%d",&n,&m)!= EOF)
{
init();
for(int i=,u,v,d,a;i<m;i++)
{
scanf("%d%d%d%d",&u,&v,&d,&a);
addEdge(u,v,,d);
addEdge(u,v,,d+a);
}
S=,T=n+;
addEdge(S,,,);
addEdge(n,T,,);
int cost,flow;
MCMF(S,T,cost,flow);
printf("Case %d: %d\n",++ca,cost);
}
return ;
}