[ACdream 1211 Reactor Cooling]无源无汇有上下界的可行流

题意:无源无汇有上下界的可行流 模型

思路:首先将所有边的容量设为上界减去下界,然后对一个点i,设i的所有入边的下界和为to[i],所有出边的下界和为from[i],令它们的差为dif[i]=to[i]-from[i],根据流量平衡原理,让出边和入边的下界相抵消,如果dif[i]>0,说明入边把出边的下界抵消了,还剩下dif[i]的流量必须要流过来(否则不满足入边的下界条件),这时从源点向i连一条容量为dif[i]的边来表示即可,如果dif[i]<0,同理应该从i向汇点连一条容量为-dif[i]的边。最后对新建好的图跑一遍最大流,如果源点的所有出边都满流了说明原图有可行流,可行解为每条边在新图的流量加上它的下界。

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#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; #define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull; #ifndef ONLINE_JUDGE
namespace Debug {
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
#endif // ONLINE_JUDGE template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double EPS = 1e-14; /* -------------------------------------------------------------------------------- */ const int maxn = 2e2 + ; struct Dinic {
private:
//const static int maxn = 1e3 + 7;
struct Edge {
int from, to, cap, least;
Edge(int u, int v, int w, int l): from(u), to(v), cap(w), least(l) {}
};
int s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn], cur[maxn]; bool bfs() {
memset(vis, , sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = ;
vis[s] = true;
while (!Q.empty()) {
int x = Q.front(); Q.pop();
for (int i = ; i < G[x].size(); i ++) {
Edge &e = edges[G[x][i]];
if (!vis[e.to] && e.cap) {
vis[e.to] = true;
d[e.to] = d[x] + ;
Q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x, int a) {
if (x == t || a == ) return a;
int flow = , f;
for (int &i = cur[x]; i < G[x].size(); i ++) {
Edge &e = edges[G[x][i]];
if (d[x] + == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > ) {
e.cap -= f;
edges[G[x][i] ^ ].cap += f;
flow += f;
a -= f;
if (a == ) break;
}
}
return flow;
} public:
void clear() {
for (int i = ; i < maxn; i ++) G[i].clear();
edges.clear();
memset(d, , sizeof(d));
}
void add(int from, int to, int cap, int least) {
edges.push_back(Edge(from, to, cap, least));
edges.push_back(Edge(to, from, , least));
int m = edges.size();
G[from].push_back(m - );
G[to].push_back(m - );
} int solve(int s, int t) {
this->s = s; this->t = t;
int flow = ;
while (bfs()) {
memset(cur, , sizeof(cur));
flow += dfs(s, 1e9);
}
return flow;
} void out(int m) {
for (int i = ; i < m; i ++) {
printf("%d\n", edges[i << ].least + edges[i << | ].cap);
}
}
};
Dinic solver;
int tob[maxn], fromb[maxn]; int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int n, m;
while (cin >> n >> m) {
solver.clear();
fillchar(tob, );
fillchar(fromb, );
for (int i = ; i < m; i ++) {
int u, v, b, c;
scanf("%d%d%d%d", &u, &v, &b, &c);
solver.add(u, v, c - b, b);
tob[v] += b;
fromb[u] += b;
}
int total = ;
for (int i = ; i <= n; i ++) {
int dif = tob[i] - fromb[i];
if (dif > ) solver.add(, i, dif, );
if (dif < ) solver.add(i, n + , - dif, );
total += abs(dif);
}
if (solver.solve(, n + ) != total / ) puts("NO");
else {
puts("YES");
solver.out(m);
}
}
return ;
}
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