Crazy Circuits
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 611 Accepted Submission(s): 308
Problem Description
You’ve just built a circuit board for your new robot, and now you need to power it. Your robot circuit consists of a number of electrical components that each require a certain amount of current to operate. Every component has a + and a - lead, which are connected
on the circuit board at junctions. Current flows through the component from + to - (but note that a component does not "use up" the current: everything that comes in through the + end goes out the - end).
The junctions on the board are labeled 1, ..., N, except for two special junctions labeled + and - where the power supply terminals are connected. The + terminal only connects + leads, and the - terminal only connects - leads. All current that
enters a junction from the - leads of connected components exits through connected + leads, but you are able to control how much current flows to each connected + lead at every junction (though methods for doing so are beyond the scope of this problem1).
Moreover, you know you have assembled the circuit in such a way that there are no feedback loops (components chained in a manner that allows current to flow in a loop).
Figure 1: Examples of two valid circuit diagrams.
In (a), all components can be powered along directed paths from the positive terminal to the negative terminal.
In (b), components 4 and 6 cannot be powered, since there is no directed path from junction 4 to the negative terminal.
In the interest of saving power, and also to ensure that your circuit does not overheat, you would like to use as little current as possible to get your robot to work. What is the smallest amount of current that you need to put through the + terminal (which
you can imagine all necessarily leaving through the - terminal) so that every component on your robot receives its required supply of current to function?
Hint
1 For those who are electronics-inclined, imagine that you have the ability to adjust the potential on any componentwithout altering its current requirement, or equivalently that there is an accurate variable potentiometer connected in series with
each component that you can adjust. Your power supply will have ample potential for the circuit.
Input
The input file will contain multiple test cases. Each test case begins with a single line containing two integers: N (0 <= N <= 50), the number of junctions not including the positive and negative terminals, and M (1
<= M <= 200), the number of components in the circuit diagram. The next M lines each contain a description of some component in the diagram. The ith component description contains three fields: pi,
the positive junction to which the component is connected, ni, the negative junction to which the component is connected, and an integer Ii (1 <= Ii <= 100), the minimum amount
of current required for component i to function. The junctions pi and ni are specified as either the character ‘+‘ indicating the positive terminal, the character ‘-‘ indicating the negative
terminal, or an integer (between 1 and N) indicating one of the numbered junctions. No two components have the same positive junction and the same negative junction. The end-of-file is denoted by an invalid test case with N =M =
0 and should not be processed.
Output
For each input test case, your program should print out either a single integer indicating the minimum amount of current that must be supplied at the positive terminal in order to ensure that every component is powered, or the message "impossible"
if there is no way to direct a sufficient amount of current to each component simultaneously.
Sample Input
6 10 + 1 1 1 2 1 1 3 2 2 4 5 + - 1 4 3 2 3 5 5 4 6 2 5 - 1 6 5 3 4 6 + 1 8 1 2 4 1 3 5 2 4 6 3 - 1 3 4 3 0 0
Sample Output
9 impossible
Source
2008 ACM-ICPC Pacific Northwest Region
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题意：有两个正负极n个节点和m个元件，每个元件告诉端点是接在哪个节点上的，并且每个元件有工作的最小电流限制，问使所有元件工作的满足条件的最小电流是多少。
思路：题目中已经有了源点S和汇点T，再添加附加源点SS和汇点TT，原图中的边的容量为INF（因为没有上限），若in[i]>0则SS向i连边容量为in[i]，若in[i]<0则i向TT连边容量为-in[i]，跑一次最大流，然后T向S连容量为INF的边，再跑一次最大流，判断SS的出边是否满流，若不满流则无解，否则有解输出边(T,S)的流量。
另外这类题看这里详解：点击打开链接
代码：
#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
int n,m;
char s1[10],s2[10];
int in[MAXN];
int S,T,SS,TT;
void init()
{
tol=0;
memset(head,-1,sizeof(head));
memset(in,0,sizeof(in));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
edge[tol].flow=0; head[u]=tol++;
edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
edge[tol].flow=0; head[v]=tol++;
}
//输入参数：起点，终点，点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
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!=-1;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!=-1;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!=-1;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=head[u];i!=-1;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;
}
void solve()
{
for (int i=0;i<=n+1;i++)
{
if (in[i]>0) addedge(SS,i,in[i]);
if (in[i]<0) addedge(i,TT,-in[i]);
}
sap(SS,TT,n+4);
addedge(T,S,INF);
sap(SS,TT,n+4);
for (int i=head[SS];~i;i=edge[i].next)
if (edge[i].cap-edge[i].flow>0)
{
printf("impossible\n");
return ;
}
printf("%d\n",edge[head[T]].flow);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v,w;
while (scanf("%d%d",&n,&m)&&(n+m))
{
init();
S=0,T=n+1,SS=n+2,TT=n+3;
for (i=0;i<m;i++)
{
scanf("%s%s%d",s1,s2,&w);
if (s1[0]=='+') u=S;
else
{
u=0;
for (j=0;j<strlen(s1);j++)
u=u*10+s1[j]-'0';
}
if (s2[0]=='-') v=T;
else
{
v=0;
for (j=0;j<strlen(s2);j++)
v=v*10+s2[j]-'0';
}
in[v]+=w;
in[u]-=w;
addedge(u,v,INF);
}
solve();
}
return 0;
}
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