BZOJ 1061费用流

2019-11-06 07:29:46

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思路：我们可以列出几个不等式用y0带进去变成等式下-上可以消好多东西

我们发现等式左边的加起来=0 可以把每个方程看成一个点

正->负连边跑费用流即可

//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))const int N=24005,M=1005,inf=0x3f3f3f3f;int edge[N],cost[N],v[N],w[N],first[M],next[N],tot,ans;int n,m,xx,yy,zz,least[M],with[M],vis[N],d[M],minn[M];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(with,0),mem(vis,0),mem(d,0x3f),mem(minn,0x3f); queue<int>q;q.push(0);d[0]=0; while(!q.empty()){ int t=q.front();q.pop();vis[t]=0; for(int i=first[t];~i;i=next[i]){ if(d[v[i]]>d[t]+cost[i]&&edge[i]){ d[v[i]]=d[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 d[n+1]<inf;}int zeng(){ for(int i=n+1;i;i=v[with[i]^1]) edge[with[i]]-=minn[n+1],edge[with[i]^1]+=minn[n+1]; return minn[n+1]*d[n+1];}signed main(){ mem(first,-1); scanf("%lld%lld",&n,&m); for(int i=1;i<=n;i++){ scanf("%lld",&least[i]); if(least[i]-least[i-1]>0)add(0,i,0,least[i]-least[i-1]); else add(i,n+1,0,least[i-1]-least[i]); if(i!=n)add(i+1,i,0,inf); } for(int i=1;i<=m;i++) scanf("%lld%lld%lld",&xx,&yy,&zz),add(xx,yy+1,zz,inf); while(tell())ans+=zeng(); PRintf("%lld/n",ans);}

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