题意:一个电路板,上面有N个接线柱(标号1~N) 还有两个电源接线柱 + - 然后是 给出M个部件正负极的接线柱和最小电流,求一个可以让所有部件正常工作的总电流。
析:这是一个有源汇有上下界的最小流。
有源汇有上下界最大流:
1.构造附加网络
2.对ss、tt求最大流(ss、tt满流则有解)
3.若有解,对s、t求最大流
有源汇有上下界最小流:
1.构造附加网络(不添加[t,s]边)
2.对ss、tt求最大流
3.添加[t,s]边
4.对ss、tt求最大流
5.若ss、tt满流,则[t,s]的流量就是最小流
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 80 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
} struct Edge{
int from, to, cap, flow;
}; struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn]; void init(int n){
this-> n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
} void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m-2); G[to].pb(m-1);
} bool bfs(){
ms(vis, 0); vis[s] = 1;
d[s] = 1;
queue<int> q;
q.push(s); while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
d[e.to] = d[x] + 1;
vis[e.to] = 1;
q.push(e.to);
}
}
}
return vis[t];
} int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
a -= f;
flow += f;
if(a == 0) break;
}
}
return flow;
} int maxflow(int s, int t){
this->s = s; this-> t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
} void solve(int t){
for(int i = 0; i < G[s].sz; ++i){
Edge &e = edges[G[s][i]];
if(e.cap > e.flow){
puts("impossible");
return ;
}
}
printf("%d\n", edges[*G[t].rbegin()].flow);
}
}; Dinic dinic;
int in[maxn]; int main(){
while(scanf("%d %d", &n, &m) == 2 && n+m){
int s = n + 2, t = n + 3;
dinic.init(n + 10);
char a[15], b[15], c[15];
ms(in, 0);
for(int i = 0; i < m; ++i){
scanf("%s %s %s", a, b, c);
int u = a[0] == '+' ? 0 : atoi(a);
int v = b[0] == '-' ? n+1 : atoi(b);
int val = atoi(c);
in[v] += val;
in[u] -= val;
dinic.addEdge(u, v, INF);
}
for(int i = 0; i <= n+1; ++i){
if(in[i] > 0) dinic.addEdge(s, i, in[i]);
if(in[i] < 0) dinic.addEdge(i, t, -in[i]);
}
dinic.maxflow(s, t);
dinic.addEdge(n+1, 0, INF);
dinic.maxflow(s, t);
dinic.solve(n+1);
}
return 0;
}
Crazy Circuits