Internship
Time Limit: 5 Seconds
Memory Limit: 32768 KB
CIA headquarter collects data from across the country through its classified network. They have been usingoptical fibres long before it's been deployed on any civilian projects. However they are still under a lotpressure recently because the data are growing
rapidly. As a result they are considering upgrading thenetwork with new technologies that provide a few times wider bandwidth. In the experiemental stage,they would like to upgrade one segment of their original network in order to see how it performs. Andas
a CIA intern it's your responsibility to investigate which segment could actually help increasethe total bandwidth the headquarter receives, suppose that all the cities have infinite data to sendand the routing algorithm is optimized. As they have prepared
the data for you in a few minutes, you aretold that they need the result immediately. Well, practically immediately.
Input
Input contains multiple test cases. First line of each test case contains three integers n, m and l, theyrepresent the number of cities, the number of relay stations and the number of segments. Cities will bereferred to as integers from 1 to n, while relay
stations use integers from n+1 to n+m. You can savesassume that n + m <= 100, l <= 1000 (all of them are positive). The headquarter is identified by theinteger 0.
The next l lines hold a segment on each line in the form of a b c, where a is the source node and b isthe target node, while c is its bandwidth. They are all integers where a and b are valid identifiers(from 0 to n+m). c is positive. For some reason the
data links are all directional.
The input is terminated by a test case with n = 0. You can safely assume that your calculation canbe housed within 32-bit integers.
Output
For each test print the segment id's that meets the criteria. The result is printed in a single lineand sorted in ascending order, with a single space as the separator. If none of the segment meets thecriteria, just print an empty line. The segment id is
1 based not 0 based.
Sample Input
2 1 3
1 3 2
3 0 1
2 0 1
2 1 3
1 3 1
2 3 1
3 0 2
0 0 0
Sample Output
2 3
<hey here is an invisible empty line>
Author: WU, Jiazhi
Source: CYJJ's Funny Contest #3, Having Fun in Summer
题意：CIA公司想采用新技术升级网络，在实验测试阶段，他们想升级其中的一段网络以便观察新技术在多大的长度上提升网络的性能，你作为实习生的任务是调查那一段网络能提高CIA总部的宽带。
思路：判断一段网络可不可以提升网络就要看它是不是满流，如果满流则可能在升级后提升CIA总部的宽带，但是如果提升后并不能增广，即不能提升CIA总部的宽带，所以判断一段是不是可提升的则有两个条件：（1）在进行增广后这段网络是满流的，（2）在提升后可以增广。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; typedef struct node
{ int u; int v; int Flow; int next;
}Line; Line Li[2200]; int Head[110],top; int vis[110],ans[110]; bool vis1[110],vis2[110]; int n,m,l; int s,t; void AddEdge(int u,int v,int f)
{
Li[top].v=v; Li[top].u=u;
Li[top].Flow=f;
Li[top].next=Head[u];
Head[u]=top++;
} bool BFS()
{
memset(vis,-1,sizeof(vis)); vis[s]=0; queue<int >Q; Q.push(s); while(!Q.empty())
{ int u=Q.front(); Q.pop(); for(int i=Head[u];i!=-1;i=Li[i].next)
{ if(Li[i].Flow&&vis[Li[i].v]==-1)
{
vis[Li[i].v]=vis[u]+1; Q.push(Li[i].v); }
}
}
return vis[t]!=-1; } int DFS(int u,int f)
{
if(u==t)
{
return f;
}
int ans=0; for(int i=Head[u];i!=-1;i=Li[i].next)
{
if(Li[i].Flow&&vis[Li[i].v]==vis[u]+1)
{
int d=DFS(Li[i].v,min(f,Li[i].Flow));
f-=d;
Li[i].Flow-=d;
Li[i^1].Flow+=d;
ans+=d;
}
}
return ans;
} void dfs(int u,bool *vist,int op)
{
vist[u]=true; for(int i=Head[u];i!=-1;i=Li[i].next)
{
if(!vist[Li[i].v]&&Li[i^op].Flow!=0)
{
dfs(Li[i].v,vist,op);
}
}
} void Dinic()//网络流进行增广
{
int ans; while(BFS())
{
ans=DFS(s,INF); }
} int main()
{
while(~scanf("%d %d %d",&n,&m,&l))
{ if(n+m+l==0)
{
break;
} s=n+m+1;//源点 t=0;//汇点 memset(Head,-1,sizeof(Head)); int a,b,c; top = 0; for(int i=0;i<l;i++)
{
scanf("%d %d %d",&a,&b,&c);
AddEdge(a,b,c);//建立边，正向为c，负向为0
AddEdge(b,a,0);
} for(int i=1;i<=n;i++)
{
AddEdge(s,i,INF); AddEdge(i,s,0);//建立城市与源点之间的边，权值为INF
} Dinic(); memset(vis1,false,sizeof(vis1)); memset(vis2,false,sizeof(vis2)); dfs(s,vis1,0);//从源点向汇点搜索，标记还有剩余流的点 dfs(t,vis2,1);//从汇点到源点搜索，标记还有剩余流的点 int num=0; for(int i=0;i<l;i++)
{
if(Li[i<<1].Flow==0&&vis1[Li[i<<1].u]&&vis2[Li[i<<1].v])
{
ans[num++]=i+1;//如果一条边的u与v都被标记，表明s->u,v->t,但是这条边是满流，所以提升这条边。
}
} if(num)
{
for(int i=0;i<num;i++)
{
if(i)
{
printf(" ");
}
printf("%d",ans[i]);
}
}
printf("\n"); }
return 0;
}