题意:有N个电子元件,每个电子元件需要一个最少的电能驱动,问最少需要多少电能能把所有元件驱动。

最小流做法
代码:

//author: CHC
//First Edit Time: 2015-09-10 01:48
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#include <limits>
using namespace std;
typedef long long LL;
const int MAXN=1e+4;
const int MAXM=1e+5;
const int INF = numeric_limits<int>::max();
const LL LL_INF= numeric_limits<LL>::max();
struct Edge {
int to,ci,next;
Edge(){}
Edge(int _to,int _ci,int _next):to(_to),ci(_ci),next(_next){}
}e[MAXM];
int head[MAXN],tot;
void init(){
memset(head,-1,sizeof(head));
tot=0;
}
void AddEdge(int u,int v,int ci0,int ci1=0){
e[tot]=Edge(v,ci0,head[u]);
head[u]=tot++;
e[tot]=Edge(u,ci1,head[v]);
head[v]=tot++;
}
int dis[MAXN],sta[MAXN],top,cur[MAXN];
bool bfs(int st,int et){
memset(dis,0,sizeof(dis));
dis[st]=1;
queue <int> q;
q.push(st);
while(!q.empty()){
int now=q.front(); q.pop();
for(int i=head[now];~i;i=e[i].next){
if(e[i].ci&&!dis[e[i].to]){
dis[e[i].to]=dis[now]+1;
q.push(e[i].to);
}
}
}
return dis[et]!=0;
}
LL Dinic(int st,int et){
LL ans=0;
while(bfs(st,et)){
top=0;
memcpy(cur,head,sizeof(cur));
int u=st,i;
while(1){
if(u==et){
int pos,minn=INF;
for(i=0;i<top;i++){
if(minn>e[sta[i]].ci){
minn=e[sta[i]].ci;
pos=i;
}
}
for(i=0;i<top;i++){
e[sta[i]].ci-=minn;
e[sta[i]^1].ci+=minn;
}
top=pos;
u=e[sta[top]^1].to;
ans+=minn;
}
for(i=cur[u];~i;cur[u]=i=e[i].next)
if(e[i].ci&&dis[u]+1==dis[e[i].to])break;
if(cur[u]!=-1){
sta[top++]=cur[u];
u=e[cur[u]].to;
}
else {
if(top==0)break;
dis[u]=0;
u=e[sta[--top]^1].to;
}
}
}
return ans;
}
int du[MAXN],n,m;
char str1[10],str2[10];
int main()
{
while(~scanf("%d%d",&n,&m)){
if(n+m==0)break;
init();
memset(du,0,sizeof(du));
int st=0,et=n+1;
for(int i=0,x,y,v;i<m;i++){
scanf("%s%s%d",str1,str2,&v);
if(str1[0]=='+') x=st;
else sscanf(str1,"%d",&x);
if(str2[0]=='-') y=et;
else sscanf(str2,"%d",&y);
AddEdge(x,y,INF-v);
du[x]-=v;du[y]+=v;
}
int sst=et+1,eet=et+2;
LL ctot=0;
for(int i=0;i<=et;i++){
if(du[i]>0)AddEdge(sst,i,du[i]),ctot+=du[i];
else AddEdge(i,eet,-du[i]);
}
LL ans=Dinic(sst,eet);
AddEdge(et,st,INF);
ans+=Dinic(sst,eet);
if(ans!=ctot)puts("impossible");
else printf("%d\n",e[tot-1].ci);
}
return 0;
}