POJ3068："Shortest" pair of paths——题解

http://poj.org/problem?id=3068

题目大意：

从0～n-1找到两条边和点都不相同（除了0和n-1外）的最小费用路径。

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POJ2135魔改版。

按照那题的思路并且把点拆成中间连一条容量为1的边即可。

切了切了。

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
typedef long long ll;
const int INF=1e9;
const int N=70;
const int M=20100;
inline int read(){
    int X=0,w=0;char ch=0;
    while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
    while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
struct node{
    int nxt;
    int to;
    int w;
    int b;
}edge[M];
int head[N],cnt=-1;
void add(int u,int v,int w,int b){
    cnt++;
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].b=b;
    edge[cnt].nxt=head[u];
    head[u]=cnt;
    return;
}
int dis[N];
bool vis[N];
inline bool spfa(int s,int t,int n){
    deque<int>q;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)dis[i]=INF;
    dis[t]=0;q.push_back(t);vis[t]=1;
    while(!q.empty()){
    int u=q.front();
    q.pop_front();vis[u]=0;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        int b=edge[i].b;
        if(edge[i^1].w&&dis[v]>dis[u]-b){
        dis[v]=dis[u]-b;
        if(!vis[v]){
            vis[v]=1;
            if(!q.empty()&&dis[v]<dis[q.front()]){
            q.push_front(v);
            }else{
            q.push_back(v);
            }
        }
        }
    }
    }
    return dis[s]<INF;
}
int ans=0;
int dfs(int u,int flow,int m){
    if(u==m){
    vis[m]=1;
    return flow;
    }
    int res=0,delta;
    vis[u]=1;
    for(int e=head[u];e!=-1;e=edge[e].nxt){
        int v=edge[e].to;
    int b=edge[e].b;
        if(!vis[v]&&edge[e].w&&dis[u]-b==dis[v]){
            delta=dfs(v,min(edge[e].w,flow-res),m);
            if(delta){
                edge[e].w-=delta;
                edge[e^1].w+=delta;
                res+=delta;
        ans+=delta*b;
                if(res==flow)break;
            }
        }
    }
    return res;
}
inline int costflow(int S,int T,int n){
    int flow=0;
    while(spfa(S,T,n)){
    do{
        memset(vis,0,sizeof(vis));
        flow+=dfs(S,INF,T);
    }while(vis[T]);
    }
    return flow;
}
void restart(){
    memset(head,-1,sizeof(head));
    cnt=-1;
    ans=0;
    return;
}
int main(){
    int ecnt=0,n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&n&&m){
    restart();
    ecnt++;
    int F=1,E=n;
    int S=n*2+1,T=S+1;
    add(S,F,2,0);add(F,S,0,0);
    add(E,T,2,0);add(T,E,0,0);
    for(int i=2;i<=n-1;i++){
        add(i,i+n,1,0);
        add(i+n,i,0,0);
    }
    for(int i=1;i<=m;i++){
        int u=read()+1;
        int v=read()+1;
        int b=read();
        if(u!=1&&u!=n)u+=n;
        add(u,v,1,b);
        add(v,u,0,-b);
    }
    int flow=costflow(S,T,T);
    printf("Instance #%d: ",ecnt);
    if(flow!=2){
        printf("Not possible\n");
    }else printf("%d\n",ans);
    }
    return 0;
}