ZOJ 2314 Reactor Cooling

Reactor Cooling

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 2314
64-bit integer IO format: %lld      Java class name: Main

Special Judge

 

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 fij, (put fij = 0 if there is no pipe from node i to node j), for each i the following condition must hold:

fi,1+fi,2+...+fi,N = f1,i+f2,i+...+fN,i

Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be fij <= cij where cij 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 lij, thus it must be fij >= lij.

Given cij and lij for all pipes, find the amount fij, satisfying the conditions specified above.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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, lij and cij each. There is at most one pipe connecting any two nodes and 0 <= lij <= cij <= 10^5 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.

Sample Input

2

4 6
1 2 1 2
2 3 1 2
3 4 1 2
4 1 1 2
1 3 1 2
4 2 1 2

4 6
1 2 1 3
2 3 1 3
3 4 1 3
4 1 1 3
1 3 1 3
4 2 1 3

Sample Input

NO

YES
1
2
3
2
1
1

 

Source

Andrew Stankevich's Contest #1

 

解题：无源汇的上下界流。建图是关键。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct arc {
     int to,flow,next;
     arc(int x = ,int y = ,int z = -) {
         to = x;
         flow = y;
         next = z;
     }
 };
 arc e[maxn*maxn];
 int head[maxn],d[maxn],cur[maxn],tot,S,T;
 int q[maxn<<],hd,tl;
 int limit[maxn],b[];
 void add(int u,int v,int w) {
     e[tot] = arc(v,w,head[u]);
     head[u] = tot++;
     e[tot] = arc(u,,head[v]);
     head[v] = tot++;
 }
 bool bfs() {
     memset(d,-,sizeof(d));
     hd = tl = ;
     q[tl++] = S;
     d[S] = ;
     while(hd < tl) {
         int u = q[hd++];
         for(int i = head[u]; ~i; i = e[i].next) {
             if(e[i].flow && d[e[i].to] == -) {
                 d[e[i].to] = d[u] + ;
                 q[tl++] = e[i].to;
             }
         }
     }
     return d[T] > -;
 }
 int dfs(int u,int low) {
     if(u == T) return low;
     int tmp = ,a;
     for(int &i = cur[u]; ~i; i = e[i].next) {
         if(e[i].flow && d[e[i].to] == d[u] +  && (a = dfs(e[i].to,min(e[i].flow,low)))) {
             tmp += a;
             low -= a;
             e[i].flow -= a;
             e[i^].flow += a;
             if(!low) break;
         }
     }
     if(!tmp) d[u] = -;
     return tmp;
 }
 int dinic(){
     int flow = ;
     while(bfs()){
         memcpy(cur,head,sizeof(head));
         flow += dfs(S,INF);
     }
     return flow;
 }
 int main() {
     int u,v,c,w,n,m,cs;
     scanf("%d",&cs);
     while(cs--){
         scanf("%d %d",&n,&m);
         memset(head,-,sizeof(head));
         memset(limit,,sizeof(limit));
         S = tot = ;
         T = n + ;
         for(int i = ; i < m; i++){
             scanf("%d %d %d %d",&u,&v,b+i,&c);
             add(u,v,c - b[i]);
             limit[u] -= b[i];
             limit[v] += b[i];
         }
         int fullflow = ;
         for(int i = ; i <= n; i++)
             if(limit[i] > ){
                 add(S,i,limit[i]);
                 fullflow += limit[i];
             }else if(limit[i] < ) add(i,T,-limit[i]);
         w = dinic();
         if(w != fullflow) puts("NO");
         else {
             puts("YES");
             for(int i = ; i < m; i++)
                 printf("%d\n",e[i<<^].flow+b[i]);
         }
         if(cs) puts("");
     }
     return ;
 }

另外一种建图方法

 #include <bits/stdc++.h>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 const int maxn = ;
 struct arc{
     int to,flow,next;
     arc(int x = ,int y = ,int z = -){
         to = x;
         flow = y;
         next = z;
     }
 }e[maxn*maxn];
 int head[maxn],d[maxn],cur[maxn],S,T,tot;
 void add(int u,int v,int flow){
     e[tot] = arc(v,flow,head[u]);
     head[u] = tot++;
     e[tot] = arc(u,,head[v]);
     head[v] = tot++;
 }
 bool bfs(){
     queue<int>q;
     memset(d,-,sizeof d);
     d[S] = ;
     q.push(S);
     while(!q.empty()){
         int u = q.front();
         q.pop();
         for(int i = head[u]; ~i; i = e[i].next){
             if(e[i].flow && d[e[i].to] == -){
                 d[e[i].to] = d[u] + ;
                 q.push(e[i].to);
             }
         }
     }
     return d[T] > -;
 }
 int dfs(int u,int low){
     if(u == T) return low;
     int a,tmp = ;
     for(int &i = cur[u]; ~i; i = e[i].next){
         if(e[i].flow && d[e[i].to] == d[u] + &&(a=dfs(e[i].to,min(e[i].flow,low)))){
             e[i].flow -= a;
             e[i^].flow += a;
             low -= a;
             tmp += a;
             if(!low) break;
         }
     }
     if(!tmp) d[u] = -;
     return tmp;
 }
 int dinic(int ret = ){
     while(bfs()){
         memcpy(cur,head,sizeof head);
         ret += dfs(S,INF);
     }
     return ret;
 }
 int bound[maxn*maxn];
 int main(){
     int kase,n,m,u,v,w,sum;
     scanf("%d",&kase);
     while(kase--){
         scanf("%d%d",&n,&m);
         memset(head,-,sizeof head);
         int sum = S = ;
         T = n + ;
         for(int i = tot = ; i < m; ++i){
             scanf("%d%d%d%d",&u,&v,bound + i,&w);
             add(u,v,w - bound[i]);
             add(S,v,bound[i]);
             add(u,T,bound[i]);
             sum += bound[i];
         }
         if(dinic() < sum) puts("NO");
         else{
             puts("YES");
             for(int i = ; i < m; ++i)
                 printf("%d\n",e[i*+].flow + bound[i]);
         }
     }
     return ;
 }