【网络流】 SGU 194 Reactor Cooling

无源无汇上下界网络流基础题。。。。先把上下界转化成一般的网络流形式，然后用流量平衡建图即可。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 405
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1,L,mid 
#define rson o<<1 | 1,mid+1,R
#define pii pair<int,int>
#pragma comment(linker,"/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a,LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct Edge
{
	int v,c,next;
	Edge() {}
	Edge(int v,int c,int next) : v(v),c(c),next(next) {} 
}E[maxm];

queue<int> q;
int H[maxn],cntE;
int dis[maxn];
int cur[maxn];
int cnt[maxn];
int pre[maxn];
int in[maxn];
int du[maxn];
int out[maxn];
int pipes[maxm];
int flow,s,t,nv,n,m,res;

void addedges(int u,int v,int c)
{
	E[cntE] = Edge(v,H[u]);
	H[u] = cntE++;
	E[cntE] = Edge(u,H[v]);
	H[v] = cntE++;
}

void bfs()
{
	memset(cnt,sizeof cnt);
	memset(dis,-1,sizeof dis);
	cnt[0] = 1,dis[t] = 0;
	q.push(t);
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		for(int e = H[u]; ~e; e = E[e].next) {
			int v = E[e].v;
			if(dis[v] == -1) {
				dis[v] = dis[u] + 1;
				cnt[dis[v]]++;
				q.push(v);
			}
		}
	}
}

int isap()
{
	memcpy(cur,H,sizeof cur);
	flow = 0;
	bfs();
	int u = pre[s] = s,e,minv,f,pos;
	while(dis[s] < nv) {
		if(u == t) {
			f = INF;
			for(int i = s; i != t; i = E[cur[i]].v) if(E[cur[i]].c < f) {
				f = E[cur[i]].c;
				pos = i;
			}
			for(int i = s; i != t; i = E[cur[i]].v) {
				E[cur[i]].c -= f;
				E[cur[i] ^ 1].c += f;
			}
			flow += f;
			u = pos;
		}
		for(e = H[u]; ~e; e = E[e].next) if(E[e].c && dis[E[e].v] + 1 == dis[u]) break;
		if(~e) {
			cur[u] = e;
			pre[E[e].v] = u;
			u = E[e].v;
		}
		else {
			if(--cnt[dis[u]] == 0) break;
			for(e = H[u],minv = nv; ~e; e = E[e].next) if(E[e].c && minv > dis[E[e].v]) {
				minv = dis[E[e].v];
				cur[u] = e;
			}
			dis[u] = minv + 1;
			cnt[dis[u]]++;
			u = pre[u];
		}
	}
	return flow;
}

void init()
{
	res = cntE = 0;
	memset(H,sizeof H);
	memset(in,sizeof in);
	memset(du,sizeof du);
	memset(out,sizeof out);
}

void read()
{
	int u,v,a,b;
	for(int i = 0; i < m; i++) {
		scanf("%d%d%d%d",&u,&v,&a,&b);
		int c = b - a;
		in[v] += a;
		out[u] += a;
		pipes[i] = a;
		addedges(u,c);
	}
}

void work()
{
	s = 0,t = n + 1,nv = t + 1;
	for(int i = 1; i <= n; i++) {
		du[i] = in[i] - out[i];
		if(du[i] > 0) addedges(s,i,du[i]),res += du[i];
		else addedges(i,-du[i]);
	}
	if(isap() == res) {
		printf("YESn");
		for(int i = 0; i < m; i++) printf("%dn",E[i * 2 + 1].c + pipes[i]);
	}
	else printf("NOn");
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF) {
		init();
		read();
		work();
	}
	
	
	return 0;
}

（编辑：李大同）

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