SGU 194 Reactor Cooling（无源无汇上下界可行流）

Description

The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group, you are responsible for developing the cooling system for the reactor. 
The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points, called nodes, each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction. 
Let the nodes be numbered from 1 to N. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is, if we designate the amount of liquid going by the pipe from i-th node to j-th as f_ij, (put f_ij = 0 if there is no pipe from node i to node j), for each i the following condition must hold: sum(j=1..N, f_ij) = sum(j=1..N, f_ji) Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be f_ij ≤ c_ij where c_ij is the capacity of the pipe. To provide sufficient cooling, the amount of the liquid flowing by the pipe going from i-th to j-th nodes must be at least l_ij, thus it must be f_ij ≥ l_ij. 
Given c_ij and l_ij for all pipes, find the amount f_ij, satisfying the conditions specified above. 

Input

The first line of the input file contains the number N (1 ≤ N ≤ 200) - the number of nodes and and M — the number of pipes. The following M lines contain four integer number each - i, j, l_ij and c_ijeach. There is at most one pipe connecting any two nodes and 0 ≤ l_ij ≤ c_ij ≤ 10⁵ for all pipes. No pipe connects a node to itself. If there is a pipe from i-th node to j-th, there is no pipe from j-th node to i-th. 

Output

On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case M integers must follow, k-th number being the amount of liquid flowing by the k-th pipe. Pipes are numbered as they are given in the input file. 

 

题目大意：用n个点，m条有向边，每条边有一个容量的上下界，求一个可行流，要求每个点的入流等于出流。

思路：记f[i] = ∑(u,i) - ∑(i,v)，其中∑(u,i)为进入i的所有边的容量下界之和，∑(i,v)为离开i的所有边的容量下界之和。建立源点S汇点T，若f[i] ≥ 0，建一条边S→i，容量为f[i]；若f[i] < 0，建一条边i→T，容量为f[i]的绝对值。对每一条边i→j，建一条边i→j，容量为上界减去下界。若最大流能使与S关联的边和与T关联的边都满流，则存在可行流，其中每条边的流量为其下界加上最大流图中的流量，否则不存在可行流。

小证明：上面的构图法乍看之下不知道为什么是对的，网上数学证明一大堆我就不说了（虽然都一样），现在我讲一种比较直观的理解。

对每一条边a→b，容量上界为up，下界为down。从S建一条边到b，容量为down；从a建一条边到T，容量为down；从a到b建一条边，容量为up-down。这样建图，若与S→b，a→T的流量都是满的，那么在原图中，我们就可以把S→b，a→T的流量换成是a→b的流量（a有down的流出，b有down的流入，满足把a有的流出，b有的流入放入边a→b，就满足了边的下界）。

之后，若对每一条边的两个点都建边到源点汇点太浪费了，所以源点S到某点i的边可以合起来，容量为∑(u,i)；同样，某点i到汇点T的边也可以合起来，容量为∑(i,v)；那么对每一个点i，都有从源点到i的边，从i到汇点的边，因为这两条边直接相连，我们只需要像上面构图所说的方法一样，保留一条就可以了。

 

代码（15MS）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int MAXN = 210;
const int MAXE = MAXN * MAXN;
const int INF = 0x3fff3fff;

struct SAP {
    int head[MAXN], gap[MAXN], dis[MAXN], cur[MAXN], pre[MAXN];
    int to[MAXE], next[MAXE], flow[MAXE], cap[MAXE];
    int n, ecnt, st, ed;

    void init() {
        memset(head, 0, sizeof(head));
        ecnt = 2;
    }

    void add_edge(int u, int v, int c) {
        to[ecnt] = v; cap[ecnt] = c; flow[ecnt] = 0; next[ecnt] = head[u]; head[u] = ecnt++;
        to[ecnt] = u; cap[ecnt] = 0; flow[ecnt] = 0; next[ecnt] = head[v]; head[v] = ecnt++;
        //printf("%d->%d %d\n", u, v, c);
    }

    void bfs() {
        memset(dis, 0x3f, sizeof(dis));
        queue<int> que; que.push(ed);
        dis[ed] = 0;
        while(!que.empty()) {
            int u = que.front(); que.pop();
            ++gap[dis[u]];
            for(int p = head[u]; p; p = next[p]) {
                int &v = to[p];
                if(cap[p ^ 1] && dis[v] > n) {
                    dis[v] = dis[u] + 1;
                    que.push(v);
                }
            }
        }
    }

    int Max_flow(int ss, int tt, int nn) {
        st = ss, ed = tt, n = nn;
        int ans = 0, minFlow = INF, u;
        for(int i = 0; i <= n; ++i) {
            cur[i] = head[i];
            gap[i] = 0;
        }
        u = pre[st] = st;
        bfs();
        while(dis[st] < n) {
            bool flag = false;
            for(int &p = cur[u]; p; p = next[p]) {
                int &v = to[p];
                if(cap[p] > flow[p] && dis[u] == dis[v] + 1) {
                    flag = true;
                    minFlow = min(minFlow, cap[p] - flow[p]);
                    pre[v] = u;
                    u = v;
                    if(u == ed) {
                        ans += minFlow;
                        while(u != st) {
                            u = pre[u];
                            flow[cur[u]] += minFlow;
                            flow[cur[u] ^ 1] -= minFlow;
                        }
                        minFlow = INF;
                    }
                    break;
                }
            }
            if(flag) continue;
            int minDis = n - 1;
            for(int p = head[u]; p; p = next[p]) {
                int &v = to[p];
                if(cap[p] > flow[p] && dis[v] < minDis) {
                    minDis = dis[v];
                    cur[u] = p;
                }
            }
            if(--gap[dis[u]] == 0) break;
            ++gap[dis[u] = minDis + 1];
            u = pre[u];
        }
        return ans;
    }
} G;

int n, m;
int f[MAXN];
int m_id[MAXE], m_down[MAXE];

int main() {
    scanf("%d%d", &n, &m);
    G.init();
    int a, b, c, d, sum = 0;
    for(int i = 1; i <= m; ++i) {
        scanf("%d%d%d%d", &a, &b, &d, &c);
        f[a] -= d;
        f[b] += d;
        m_down[i] = d;
        m_id[i] = G.ecnt;
        G.add_edge(a, b, c - d);
    }
    int ss = n + 1, tt = n + 2;
    for(int i = 1; i <= n; ++i) {
        if(f[i] >= 0) G.add_edge(ss, i, f[i]), sum += f[i];
        else G.add_edge(i, tt, -f[i]);
    }
    if(G.Max_flow(ss, tt, tt) != sum) {
        puts("NO");
        return 0;
    }
    puts("YES");
    for(int i = 1; i <= m; ++i) printf("%d\n", m_down[i] + G.flow[m_id[i]]);
}