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 666666
int 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 666666
int 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);
}