# Reactor Cooling

 Time Limit: 500MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Description

#### 194. Reactor Cooling

time limit per test: 0.5 sec.
memory limit per test: 65536 KB
input: standard
output: standard

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 ij each. There is at most one pipe connecting any two nodes and 0 ≤ l ij ≤ c ij ≤ 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 test(s)

Input

Test #1

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

Test #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

Output

Test #1

NO

Test #2

YES

  1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 using namespace std;
10 const int inf=1e9;
11 const int maxn=500,maxm=300000;
12 int N,M,S,T,maxflow;
13 struct Edge1{
14     int l,r;
15 }E[maxm];
16
17 vector<int> ru[maxn],chu[maxn],to1[maxn];
18 int in[maxn],out[maxn],delta[maxn],ans[maxn];
19
20 struct Edge2{
21     int to,rest,next;
22     int num;
23 }e[maxm];
25 inline void addedge(int x,int y,int z,int num){
28 }
29 int dis[maxn];
30 bool BFS(){
31     memset(dis,0,sizeof(dis));
32     static queue<int> Q;
33     while(!Q.empty()) Q.pop();
34     Q.push(S); dis[S]=1;
35     while(!Q.empty()){
36         int x=Q.front(); Q.pop();
38             int y=e[i].to;
39             if(e[i].rest&&dis[y]==0){
40                 dis[y]=dis[x]+1;
41                 Q.push(y);
42             }
43         }
44     }
45     if(dis[T]>0) return true;
46     return false;
47 }
48 int DFS(int x,int flow){
49     if(x==T) return flow;
50     int now=0,tmp;
52         int y=e[i].to;
53         if(e[i].rest&&dis[y]==dis[x]+1){
54             tmp=DFS(y,min(flow-now,e[i].rest));
55             e[i].rest-=tmp;
56             e[i^1].rest+=tmp;
57             now+=tmp;
58             if(now==flow) return now;
59         }
60     }
61     if(!now) dis[x]=0;
62     return now;
63 }
64 void dinic(){
65     while(BFS())
66         maxflow+=DFS(S,inf);
67 }
68 int main(){
69     scanf("%d%d",&N,&M);
70     S=0; T=N+1;
71     for(int i=1,u,v,l,r;i<=M;i++){
72         scanf("%d%d%d%d",&u,&v,&l,&r);
73         ru[v].push_back(i); chu[u].push_back(i); to1[u].push_back(v);
74         E[i].l=l; E[i].r=r;
75     }
76     for(int i=1;i<=N;i++){
77         for(int j=0;j<ru[i].size();j++){
78             int num=ru[i][j];
79             in[i]+=E[num].l;
80         }
81         for(int j=0;j<chu[i].size();j++){
82             int num=chu[i][j];
83             out[i]+=E[num].l;
84         }
85     }
86     for(int i=1;i<=N;i++) delta[i]=in[i]-out[i];
87
88     for(int i=1;i<=N;i++){
89         for(int j=0;j<chu[i].size();j++){
90             int v=to1[i][j],num=chu[i][j];
92         }
93     }
94     for(int i=1;i<=N;i++){
97     }
98
99     dinic();
101         if(e[i].rest>0){
102             puts("NO");
103             return 0;
104         }
105     }
106     puts("YES");
107     for(int i=2;i<M*2+1;i+=2) ans[e[i^1].num]=e[i^1].rest;
108     for(int i=1;i<=M;i++) printf("%d\n",ans[i]+E[i].l);
109     return 0;
110 }

zoj多组数据的如下

  1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 using namespace std;
10 const int inf=1e9;
11 const int maxn=5000,maxm=1000000;
12 int tot,N,M,S,T,maxflow;
13 struct Edge1{
14     int l,r;
15 }E[maxm];
16
17 vector<int> ru[maxn],chu[maxn],to1[maxn];
18 int in[maxn],out[maxn],delta[maxn],ans[maxn];
19
20 struct Edge2{
21     int to,rest,next;
22     int num;
23 }e[maxm];
25 inline void addedge(int x,int y,int z,int num){
28 }
29 int dis[maxn];
30 bool BFS(){
31     memset(dis,0,sizeof(dis));
32     static queue<int> Q;
33     while(!Q.empty()) Q.pop();
34     Q.push(S); dis[S]=1;
35     while(!Q.empty()){
36         int x=Q.front(); Q.pop();
38             int y=e[i].to;
39             if(e[i].rest&&dis[y]==0){
40                 dis[y]=dis[x]+1;
41                 Q.push(y);
42             }
43         }
44     }
45     if(dis[T]>0) return true;
46     return false;
47 }
48 int DFS(int x,int flow){
49     if(x==T) return flow;
50     int now=0,tmp;
52         int y=e[i].to;
53         if(e[i].rest&&dis[y]==dis[x]+1){
54             tmp=DFS(y,min(flow-now,e[i].rest));
55             e[i].rest-=tmp;
56             e[i^1].rest+=tmp;
57             now+=tmp;
58             if(now==flow) return now;
59         }
60     }
61     if(!now) dis[x]=0;
62     return now;
63 }
64 void dinic(){
65     while(BFS())
66         maxflow+=DFS(S,inf);
67 }
68 int main(){
69     scanf("%d",&tot);
70     while(tot--){
71         maxflow=0;
72         memset(ru,0,sizeof(ru)); memset(chu,0,sizeof(chu)); memset(to1,0,sizeof(to1));
73         memset(in,0,sizeof(in)); memset(out,0,sizeof(out));
74         memset(delta,0,sizeof(delta)); memset(ans,0,sizeof(ans));
76         for(int i=1;i<=cnt;i++) {
77             e[i].next=e[i].num=e[i].rest=e[i].rest=e[i].to=0;
78         }
79         cnt=1;
80
81         scanf("%d%d",&N,&M);
82         S=0; T=N+1;
83         for(int i=1,u,v,l,r;i<=M;i++){
84             scanf("%d%d%d%d",&u,&v,&l,&r);
85             ru[v].push_back(i); chu[u].push_back(i); to1[u].push_back(v);
86             E[i].l=l; E[i].r=r;
87         }
88         for(int i=1;i<=N;i++){
89             for(int j=0;j<ru[i].size();j++){
90                 int num=ru[i][j];
91                 in[i]+=E[num].l;
92             }
93             for(int j=0;j<chu[i].size();j++){
94                 int num=chu[i][j];
95                 out[i]+=E[num].l;
96             }
97         }
98         for(int i=1;i<=N;i++) delta[i]=in[i]-out[i];
99
100         for(int i=1;i<=N;i++){
101             for(int j=0;j<chu[i].size();j++){
102                 int v=to1[i][j],num=chu[i][j];
104             }
105         }
106         for(int i=1;i<=N;i++){
109         }
110
111         dinic();
113             if(e[i].rest>0){
114                 puts("NO");
115                 goto end1;
116             }
117         }
118         puts("YES");
119         for(int i=2;i<M*2+1;i+=2) ans[e[i^1].num]=e[i^1].rest;
120         for(int i=1;i<=M;i++) printf("%d\n",ans[i]+E[i].l);
121         end1:;
122     }
123     return 0;
124 }

posted @ 2016-02-21 15:32  CXCXCXC  阅读(205)  评论(0编辑  收藏  举报