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BZOJ 2324 (有上下界的)费用流

思路:

先跑一遍Floyd  更新的时候map[i][j]=map[i][k]+map[k][j]  k需要小于i或j

正常建边:

把所有点 拆点-> i,i+n

add(x,y,C,E)表示x->y建边 话费为C  容量为E

add(S,0,0,k)

add(i,j+n,map[i][j],1)

add(j+n,j,0,0x3f3f3f3f)->这条边下界为1

add(j,T,0,1)

这个时候我们有两种方法

1.套用有上下界的网络流

2.把add(j+n,j,0,0x3f3f3f3f)这条边拆成

add(j+n,j,-0x3f3f3f3f,1),add(j+n,j,0,0x3f3f3f3f)两条边

 

我都写了一遍   为啥时间差那么多....

//By SiriusRen#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define int long long#define mem(x,y) memset(x,y,sizeof(x))#define N 666666int map[155][155],xx,yy,zz,n,m,k,vis[N],with[N],dis[N],minn[N];int first[666],next[N],v[N],edge[N],cost[N],tot,S,T,jy,ans;void Add(int x,int y,int C,int E){edge[tot]=E,cost[tot]=C,v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y,int C,int E){Add(x,y,C,E),Add(y,x,-C,0);}bool tell(){    mem(vis,0),mem(with,0),mem(dis,0x3f),mem(minn,0x3f);    queue<int>q;q.push(S),dis[S]=0;    while(!q.empty()){        int t=q.front();q.pop(),vis[t]=0;        for(int i=first[t];~i;i=next[i])            if(dis[v[i]]>dis[t]+cost[i]&&edge[i]){                dis[v[i]]=dis[t]+cost[i],minn[v[i]]=min(minn[t],edge[i]),with[v[i]]=i;                if(!vis[v[i]])vis[v[i]]=1,q.push(v[i]);            }    }return dis[T]<=0x3f3f3f3fll;}int zeng(){    for(int i=T;i!=S;i=v[with[i]^1])edge[with[i]]-=minn[T],edge[with[i]^1]+=minn[T];    return dis[T]*minn[T];}signed main(){    mem(map,0x3f),mem(first,-1);    scanf("%lld%lld%lld",&n,&m,&k);    T=2*n+2,S=2*n+1;    for(int i=1;i<=n;i++)map[i][i]=0;    for(int i=1;i<=m;i++){        scanf("%lld%lld%lld",&xx,&yy,&zz);        map[xx][yy]=min(map[xx][yy],zz),map[yy][xx]=map[xx][yy];    }    for(int kk=0;kk<=n;kk++)        for(int i=0;i<=n;i++)            for(int j=0;j<=n;j++)                if(kk<=i||kk<=j)map[i][j]=min(map[i][j],map[i][kk]+map[kk][j]);    add(S,0,0,k);    for(int j=1;j<=n;j++)add(j+n,j,-0x3f3f3f3f,1),add(j+n,j,0,0x3f3f3f3f),add(j,T,0,1);    for(int i=0;i<=n;i++)        for(int j=i+1;j<=n;j++)            add(i,j+n,map[i][j],1);    while(tell())ans+=zeng();    printf("%lld\n",ans+0x3f3f3f3f*n);}

 

 

 

//By SiriusRen#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define int long long#define mem(x,y) memset(x,y,sizeof(x))#define N 666666int map[155][155],xx,yy,zz,n,m,k,vis[N],with[N],dis[N],minn[N];int first[666],next[N],v[N],edge[N],cost[N],tot,S,T,jy,ans;void Add(int x,int y,int C,int E){edge[tot]=E,cost[tot]=C,v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y,int C,int E){Add(x,y,C,E),Add(y,x,-C,0);}bool tell(){    mem(vis,0),mem(with,0),mem(dis,0x3f),mem(minn,0x3f);    queue<int>q;q.push(S),dis[S]=0;    while(!q.empty()){        int t=q.front();q.pop(),vis[t]=0;        for(int i=first[t];~i;i=next[i])            if(dis[v[i]]>dis[t]+cost[i]&&edge[i]){                dis[v[i]]=dis[t]+cost[i],minn[v[i]]=min(minn[t],edge[i]),with[v[i]]=i;                if(!vis[v[i]])vis[v[i]]=1,q.push(v[i]);            }    }return dis[T]<=0x3f3f3f3fll;}int zeng(){    for(int i=T;i!=S;i=v[with[i]^1])edge[with[i]]-=minn[T],edge[with[i]^1]+=minn[T];    return dis[T]*minn[T];}signed main(){    mem(map,0x3f),mem(first,-1);    scanf("%lld%lld%lld",&n,&m,&k);    T=2*n+2,S=2*n+1;    for(int i=1;i<=n;i++)map[i][i]=0;    for(int i=1;i<=m;i++){        scanf("%lld%lld%lld",&xx,&yy,&zz);        map[xx][yy]=min(map[xx][yy],zz),map[yy][xx]=map[xx][yy];    }    for(int kk=0;kk<=n;kk++)        for(int i=0;i<=n;i++)            for(int j=0;j<=n;j++)                if(kk<=i||kk<=j)map[i][j]=min(map[i][j],map[i][kk]+map[kk][j]);    add(S,0,0,k);    for(int i=1;i<=n;i++)add(S,i,0,1),add(i+n,T,0,1);    for(int i=0;i<=n;i++)        for(int j=i+1;j<=n;j++)            add(i,j+n,map[i][j],1);    while(tell())ans+=zeng();    printf("%lld\n",ans);}

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BZOJ 2324 (有上下界的)费用流