Route Redundancy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 625 Accepted Submission(s): 367
Problem Description
A
city is made up exclusively of one-way steets.each street in the city
has a capacity,which is the minimum of the capcities of the streets
along that route.
The redundancy ratio from point A to point B is
the ratio of the maximum number of cars that can get from point A to
point B in an hour using all routes simultaneously,to the maximum number
of cars thar can get from point A to point B in an hour using one
route.The minimum redundancy ratio is the number of capacity of the
single route with the laegest capacity.
Input
The
first line of input contains asingle integer P,(1<=P<=1000),which
is the number of data sets that follow.Each data set consists of
several lines and represents a directed graph with positive integer
weights.
The first line of each data set contains five apace
separatde integers.The first integer,D is the data set number. The
second integer,N(2<=N<=1000),is the number of nodes inthe graph.
The thied integer,E,(E>=1),is the number of edges in the graph. The
fourth integer,A,(0<=A<N),is the index of point A.The fifth
integer,B,(o<=B<N,A!=B),is the index of point B.
The
remaining E lines desceibe each edge. Each line contains three space
separated in tegers.The First integer,U(0<=U<N),is the index of
node U. The second integer,V(0<=v<N,V!=U),is the node V.The third
integer,W (1<=W<=1000),is th capacity (weight) of path from U to
V.
Output
For
each data set there is one line of output.It contains the date set
number(N) follow by a single space, followed by a floating-point value
which is the minimum redundancy ratio to 3 digits after the decimal
point.
Sample Input
1
1 7 11 0 6
0 1 3
0 3 3
1 2 4
2 0 3
2 3 1
2 4 2
3 4 2
3 5 6
4 1 1
4 6 1
5 6 9
Sample Output
1 1.667
题意:求解 最大流 / 图里面最大流量的那一条路 是多少??
http://www.cnblogs.com/yezekun/p/3925768.html正确题解提供者。
题解:此题正确解法不是在dfs时找增广路时更新,那样的话会出两个问题.所以需要先要预处理出最大流量的那条路.然后再求最大流。这样才是正确的。
#include<iostream> #include<cstdio> #include<cstring> #include <algorithm> #include <math.h> #include <queue> using namespace std; const int N = 1005; const int INF = 999999999; struct Edge{ int v,w,next; }edge[N*N]; int head[N]; int level[N]; int tot,max_increase; void init() { memset(head,-1,sizeof(head)); tot=0; } void addEdge(int u,int v,int w,int &k) { edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++; edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++; } int BFS(int src,int des) { queue<int>q; memset(level,0,sizeof(level)); level[src]=1; q.push(src); while(!q.empty()) { int u = q.front(); q.pop(); if(u==des) return 1; for(int k = head[u]; k!=-1; k=edge[k].next) { int v = edge[k].v; int w = edge[k].w; if(level[v]==0&&w!=0) { level[v]=level[u]+1; q.push(v); } } } return -1; } int dfs(int u,int des,int increaseRoad){ if(u==des||increaseRoad==0) { return increaseRoad; } int ret=0; for(int k=head[u];k!=-1;k=edge[k].next){ int v = edge[k].v,w=edge[k].w; if(level[v]==level[u]+1&&w!=0){ int MIN = min(increaseRoad-ret,w); w = dfs(v,des,MIN); if(w > 0) { edge[k].w -=w; edge[k^1].w+=w; ret+=w; if(ret==increaseRoad){ return ret; } } else level[v] = -1; if(increaseRoad==0) break; } } if(ret==0) level[u]=-1; return ret; } int Dinic(int src,int des) { int ans = 0; while(BFS(src,des)!=-1) ans+=dfs(src,des,INF); return ans; } int d,n,m,src,des; bool vis[N]; void dfs1(int u,int ans){ if(u==des){ max_increase = max(max_increase,ans); return ; } for(int k=head[u];k!=-1;k=edge[k].next){ int v = edge[k].v,w = edge[k].w; if(!vis[v]){ vis[v] = true; dfs1(v,min(w,ans)); ///最大流量由最小容量边决定 vis[v] = false; } } } int main() { int tcase; scanf("%d",&tcase); while(tcase--){ init(); max_increase = -1; scanf("%d%d%d%d%d",&d,&n,&m,&src,&des); for(int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); addEdge(u,v,w,tot); } memset(vis,false,sizeof(vis)); dfs1(src,99999999); ///deal int max_flow = Dinic(src,des); printf("%d %.3lf\n",d,max_flow*1.0/max_increase); } return 0; }