[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=0;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?1:-1; 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?1:-1;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 + 7; 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, 0, sizeof(vis)); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; 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] + 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].size(); i ++) { Edge &e = edges[G[x][i]]; if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0) { e.cap -= f; edges[G[x][i] ^ 1].cap += f; flow += f; a -= f; if (a == 0) break; } } return flow; } public: void clear() { for (int i = 0; i < maxn; i ++) G[i].clear(); edges.clear(); memset(d, 0, 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, 0, least)); int m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } int solve(int s, int t) { this->s = s; this->t = t; int flow = 0; while (bfs()) { memset(cur, 0, sizeof(cur)); flow += dfs(s, 1e9); } return flow; } void out(int m) { for (int i = 0; i < m; i ++) { printf("%d ", edges[i << 1].least + edges[i << 1 | 1].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, 0); fillchar(fromb, 0); for (int i = 0; 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 = 0; for (int i = 1; i <= n; i ++) { int dif = tob[i] - fromb[i]; if (dif > 0) solver.add(0, i, dif, 0); if (dif < 0) solver.add(i, n + 1, - dif, 0); total += abs(dif); } if (solver.solve(0, n + 1) != total / 2) puts("NO"); else { puts("YES"); solver.out(m); } } return 0; }