有源点汇点的上下界最大流问题。。。
建图非常easy。
。。
按题意就可以。。。
设原图 源点为 s 汇点 为 t，连一条t到s无下界上界无限大的边。。。。设两个超级源S，T，像无源汇推断可行流的问题一样，记录每一个点的in。连接到对应的超级源汇点。。
。对S，T跑一遍最大流，并检測S所连边是否满流。。。
假设不满足连可行流都没有无解。。
。否则去掉S，T点（但总点数不要边。
。。
在这里错了一下午）对s，t跑一遍最大流。得到的结果既答案。。。。
。
第一遍最大流保证了每一个点的下界流得到满足，此时的图里还有非常多自由流能够走。第二遍最大流就将这些流走满了得到的就是答案。。。
总结一下有源点汇点的上下界最大流 步骤为：
1：连接 t-->s INF，并添加S，T 像无源汇可行流一样建边，第一次最大流推断可行流
2：去掉S。T（Adj变-1） 总点数不变。第二次最大流得到答案
Gensokyo is a world which exists quietly beside ours, separated by a mystical border. It is a utopia where humans and other beings such as fairies, youkai(phantoms), and gods live peacefully together. Shameimaru Aya is a crow tengu with the ability to manipulate wind who has been in Gensokyo for over 1000 years. She runs the Bunbunmaru News - a newspaper chock-full of rumors, and owns the Bunkachou - her record of interesting observations for Bunbunmaru News articles and pictures of beautifuldanmaku(barrange) or cute girls living in Gensokyo. She is the biggest connoisseur of rumors about the girls of Gensokyo among the tengu. Her intelligence gathering abilities are the best in Gensokyo!
During the coming n days, Aya is planning to take many photos of m cute girls living in Gensokyo to write Bunbunmaru News daily and record at least Gx photos of girl x in total in the Bunkachou. At the k-th day, there are Ck targets, Tk1, Tk2, ..., TkCk. The number of photos of target Tki that Aya takes should be in range [Lki, Rki], if less, Aya cannot write an interesting article, if more, the girl will become angry and use her last spell card to attack Aya. What's more, Aya cannot take more than Dk photos at the k-th day. Under these constraints, the more photos, the better.
Aya is not good at solving this complex problem. So she comes to you, an earthling, for help.
Input
There are about 40 cases. Process to the end of file.
Each case begins with two integers 1 <= n <= 365, 1 <= m <= 1000. Then m integers, G1, G2, ..., Gm in range [0, 10000]. Then n days. Each day begins with two integer 1 <=C <= 100, 0 <= D <= 30000. Then C different targets. Each target is described by three integers, 0 <= T < m, 0 <= L <= R <= 100.
Output
For each case, first output the number of photos Aya can take, -1 if it's impossible to satisfy her needing. If there is a best strategy, output the number of photos of each girl Aya should take at each day on separate lines. The output must be in the same order as the input. If there are more than one best strategy, any one will be OK.
Output a blank line after each case.
Sample Input
2 3 12 12 12 3 18 0 3 9 1 3 9 2 3 9 3 18 0 3 9 1 3 9 2 3 9 2 3 12 12 12 3 18 0 3 9 1 3 9 2 3 9 3 18 0 0 3 1 3 6 2 6 9 2 3 12 12 12 3 15 0 3 9 1 3 9 2 3 9 3 21 0 0 3 1 3 6 2 6 12
Sample Output
36 6 6 6 6 6 6 36 9 6 3 3 6 9 -1
External Links
Author: WU, Zejun
Source: ZOJ Monthly, July 2009
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=2222;
const int maxm=1000000;
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow;
}edge[maxm];
int Size,Adj[maxn];
int gap[maxn],dep[maxn],pre[maxn],cur[maxn];
void init()
{
Size=0;
memset(Adj,-1,sizeof(Adj));
}
void addedge(int u,int v,int w,int rw=0)
{
edge[Size].to=v; edge[Size].cap=w; edge[Size].next=Adj[u];
edge[Size].flow=0; Adj[u]=Size++;
edge[Size].to=u; edge[Size].cap=rw; edge[Size].next=Adj[v];
edge[Size].flow=0; Adj[v]=Size++;
}
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,Adj,sizeof(Adj));
int u=start;
pre[u]=-1; gap[0]=N;
int ans=0;
while(dep[start]<N)
{
if(u==end)
{
int Min=INF;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
if(Min>edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];~i;i=edge[i].next)
{
v=edge[i].to;
if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag)
{
u=v;
continue;
}
int Min=N;
for(int i=Adj[u];~i;i=edge[i].next)
if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if(u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
int n,m;
int G[maxn],in[maxn];
int bian[maxm],low[maxm],bn;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
init(); bn=0;
memset(in,0,sizeof(in));
for(int i=1;i<=m;i++)
{
scanf("%d",&G[i]);
addedge(n+i,n+m+1,INF-G[i]);
in[n+i]-=G[i]; in[n+m+1]+=G[i];
}
for(int i=1;i<=n;i++)
{
int C,D;
scanf("%d%d",&C,&D);
addedge(0,i,D);
for(int j=0;j<C;j++)
{
int T,L,R;
scanf("%d%d%d",&T,&L,&R);
T++;
low[bn]=L;
bian[bn++]=Size;
addedge(i,n+T,R-L);
in[i]-=L; in[n+T]+=L;
}
}
addedge(n+m+1,0,INF);
int sum=0;
for(int i=0;i<n+m+2;i++)
{
if(in[i]<0) addedge(i,n+m+3,-in[i]);
if(in[i]>0)
{
sum+=in[i];
addedge(n+m+2,i,in[i]);
}
}
int MaxFlow=sap(n+m+2,n+m+3,n+m+4);
if(MaxFlow!=sum) puts("-1");
else
{
Adj[n+m+2]=Adj[n+m+3]=-1;
MaxFlow=sap(0,n+m+1,n+m+4);
printf("%d
",MaxFlow);
for(int i=0;i<bn;i++)
{
printf("%d
",edge[bian[i]].flow+low[i]);
}
}
putchar(10);
}
return 0;
}