Drainage Ditches
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 58538 | Accepted: 22485 |
Description
Every time it rains on Farmer John‘s fields, a pond forms over Bessie‘s favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie‘s
clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points
for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow
through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10
Sample Output
50
题意：有n条路，m个点，让你求出以1为源点，m为汇点的最大流；
思路：典型的最大流增光路的算法。当且仅当残量网络中不存在s-t有向道路（增广路）时，此时的流是从s到t的最大流；
ISAP()
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <set>
using namespace std;
const int inf=0x3f3f3f3f;
//head存储前向星的有相同起点的弧，num存储各层剩余节点数目，d存储各节点的层数，cur存储当前弧，pre存储路径，即当前节点的上一个节点，用于回溯。
int head[1010],num[1010],d[1010],cur[1010],n,m,cnt,s,t,q[100010],nv,pre[1010];
int maxint=inf;
struct node
{
int v,cap;
int next;
}edge[10010];
void add(int u,int v,int cap)
{
//加正向弧
edge[cnt].v=v;
edge[cnt].cap=cap;
edge[cnt].next=head[u];
head[u]=cnt++;
//加反向弧
edge[cnt].v=u;
edge[cnt].cap=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void bfs()//用bfs对每个节点进行分层
{
//算法执行之前需要用 BFS 初始化 d 数组，方法是从 t 到 s 逆向进行
int i,j;
memset(num,0,sizeof(num));
memset(d,-1,sizeof(d));
int f1=0,f2=0;
q[f1++]=t;//从
d[t]=0;
num[0]=1;
while(f2<=f1){
int u=q[f2++];
for(i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].v;
if(d[v]!=-1) continue;
d[v]=d[u]+1;//此节点的层数是上一节点的下一层
num[d[v]]++;//该层数的节点数+1
q[f1++]=v;
}
}
}
void isap()
{
memcpy(cur,head,sizeof(cur));
bfs();
int flow=0,u=pre[s]=s,i;
while(d[s]<nv){//如果d[s]大于nv，说明有断层
if(u==t){
int f=maxint,pos;
for(i=s;i!=t;i=edge[cur[i]].v){//找该路上的最少残余流量
if(f>edge[cur[i]].cap){
f=edge[cur[i]].cap;
pos=i;
}
}
for(i=s;i!=t;i=edge[cur[i]].v){//更新，将该路上的所有流量-最少参与流量
edge[cur[i]].cap-=f;
edge[cur[i]^1].cap+=f;
}
flow+=f;//加入总流量
u=pos;
}
for(i=cur[u];i!=-1;i=edge[i].next){
if(d[edge[i].v]+1==d[u]&&edge[i].cap){
break;
}
}
if(i!=-1){//如果找到可行增广路，更新
cur[u]=i;//更新当前弧
pre[edge[i].v]=u;//更新路径
u=edge[i].v;//回溯
}
else{
if(--num[d[u]]==0) break;//gap优化
int mind=nv;
for(i=head[u];i!=-1;i=edge[i].next){//找最小层次
if(edge[i].cap&&mind>d[edge[i].v]){
cur[u]=i;
mind=d[edge[i].v];
}
}
d[u]=mind+1;
num[d[u]]++;
u=pre[u];
}
}
printf("%d\n",flow);
}
int main()
{
int i,a,b,c;
while(~scanf("%d %d",&n,&m)){
memset(head,-1,sizeof(head));
cnt=0;
while(n--){
scanf("%d %d %d",&a,&b,&c);//加边
add(a,b,c);
}
s=1;
t=m;
nv=t+1;
isap();
}
return 0;
}
EK()
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
using namespace std;
#define inf 9999999999
int flow[210][210];
int maxflow[210],father[210],vis[210];
int max_flow;
int m,i;
void EK(int s,int e)
{
queue<int >q;
int u,v;
while(1)
{
memset(maxflow,0,sizeof(maxflow));//每次寻找增广路径都将每个点的流入容量置为0；
memset(vis,0,sizeof(vis));
maxflow[s]=inf;//源点的流入量置为正无穷；
q.push(s);//将源点压入队列；
while(!q.empty())//当队列不为空
{
u=q.front();
q.pop();
for(v=s;v<=e;v++)
{
if(!vis[v]&&flow[u][v]>0)
{
vis[v]=1;
father[v]=u;//记录下他的父亲方便反向更新
q.push(v);
maxflow[v]=min(maxflow[u],flow[u][v]);//当前点的容量为父亲点容量与边流量的较小者
}
}
if(maxflow[e]>0)//如果找到了汇点并且汇点容量不为0则清空队列
{
while(!q.empty())
q.pop();
break;
}
}
if(maxflow[e]==0)//已经找不到到汇点的增光路经了，就退出整个循环
break;
for(i=e;i!=s;i=father[i])
{
flow[father[i]][i]-=maxflow[e];//正向更新
flow[i][father[i]]+=maxflow[e];//反向更新
}
max_flow+=maxflow[e];//更新最大流
}
}
int main()
{
int n;
int si,ei,ci;
while(~scanf("%d %d",&n,&m))
{
max_flow=0;//最大流初始化；
memset(flow,0,sizeof(flow));
for(i=0;i<n;i++)
{
scanf("%d %d %d",&si,&ei,&ci);
flow[si][ei]+=ci;
}
EK(1,m);
printf("%d\n",max_flow);
}
}