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传送门1

Solution

两道比较裸的题。。。

复习一下最大流和费用流的模板。

Code[bzoj 2768][JLOI 2010] 冠军调查

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}return x*f;
}
#define MN 505
#define T 500
#define S 0
#define ME 200000
#define inf 0x3f3f3f3f
struct edge{int to,w,nex;}e[ME];
int hr[MN],cur[MN],en=1;
inline void ins(int f,int t,int w)
{e[++en]=(edge){t,w,hr[f]};hr[f]=en;e[++en]=(edge){f,0,hr[t]};hr[t]=en;
}
int d[MN],q[MN],top;
bool bfs()
{memset(d,0,sizeof d);register int i,j;for(d[q[i=top=1]=S]=1;i<=top;++i)for(j=hr[q[i]];j;j=e[j].nex)if(e[j].w&&!d[e[j].to])d[q[++top]=e[j].to]=d[q[i]]+1;return d[T];
}
int dfs(int x,int f)
{if(x==T) return f;int used=0,w;for(int i=cur[x];i;i=e[i].nex){cur[x]=i;if(d[e[i].to]==d[x]+1&&e[i].w){w=dfs(e[i].to,min(e[i].w,f-used));used+=w;e[i].w-=w;e[i^1].w+=w;if(f==used) return used;}}return d[x]=-1,used;
}
int ans=0;
void solve()
{while(bfs()){memcpy(cur,hr,sizeof cur);ans+=dfs(S,inf);}
}
int main()
{register int i,x,y,n=read(),m=read();for(i=1;i<=n;++i) read()?ins(S,i,1):ins(i,T,1);while(m--) x=read(),y=read(),ins(x,y,1),ins(y,x,1);solve();return 0*printf("%d\n",ans);
}

Code[bzoj 1877][SDOI 2009]晨跑

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}return x*f;
}
#define MN 505
#define ME 41000
#define inf 0x3f3f3f3f
struct edge{int to,w,c,nex;}e[ME];
int hr[MN],en=1,S,T;
inline void ins(int f,int t,int w,int c)
{e[++en]=(edge){t,w,c,hr[f]}; hr[f]=en;e[++en]=(edge){f,0,-c,hr[t]};hr[t]=en;
}
int d[MN],q[ME<<2],l,r;
bool inq[MN],vis[MN];
bool spfa()
{memset(d,0x3f,sizeof d);d[q[l=r=ME]=T]=0;inq[T]=true;register int i;while(l<=r){int u=q[l++];inq[u]=false;for(i=hr[u];i;i=e[i].nex)if(e[i^1].w&&d[e[i].to]>d[u]-e[i].c){d[e[i].to]=d[u]-e[i].c;if(inq[e[i].to]) continue;inq[e[i].to]=true;d[e[i].to]<=d[q[l]]?q[--l]=e[i].to:q[++r]=e[i].to;}}return d[S]!=inf;
}
ll mincost=0,maxflow=0;
int dfs(int x,int f)
{vis[x]=true;    if(x==T) return f;int used=0,w;for(int i=hr[x];i;i=e[i].nex)if(e[i].w&&d[e[i].to]+e[i].c==d[x]&&!vis[e[i].to]){w=dfs(e[i].to,min(e[i].w,f-used));used+=w;e[i].w-=w;e[i^1].w+=w;mincost+=1ll*w*e[i].c;if(used==f) return f; }return used;
}
void solve()
{while(spfa()){do{memset(vis,0,sizeof vis);maxflow+=dfs(S,inf);}while(vis[T]); }
}
int main()
{register int x,y,c,N=read(),M=read(),i;for(i=1;i<=N;++i) ins(i,i+N,1,0);while(M--) x=read(),y=read(),c=read(),ins(x+N,y,1,c);S=1+N;T=N;solve();return 0*printf("%lld %lld\n",maxflow,mincost);
}

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