POJ 1273 Drainage Ditches 最大流

Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 63802		Accepted: 24628

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

裸的最大流题目。。用了kuangbin大神的两套模板都A了一边。。算熟悉一下最大流了。。。Orz kuangbin

普通版：
#include<stdio.h>
#include<string.h>

const int MAXN = 210;
int maze[MAXN][MAXN];
int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
int flow[MAXN][MAXN];
int sap(int start,int end,int nodenum){
memset(cur,0,sizeof(cur));
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
memset(flow,0,sizeof(flow));
int u=pre[start]=start,maxflow=0,aug=-1;
gap[0]=nodenum;
while(dis[start]<nodenum){
loop:
for(int v=cur[u];v<nodenum;v++)
if(maze[u][v]-flow[u][v]&&dis[u]==dis[v]+1){
if(aug==-1||aug>maze[u][v]-flow[u][v]) aug=maze[u][v]-flow[u][v];
pre[v]=u;
u=cur[u]=v;
if(v==end){
maxflow+=aug;
for(u=pre[u];v!=start;v=u,u=pre[u]){
flow[u][v]+=aug;
flow[v][u]-=aug;
}
aug=-1;
}
goto loop;
}
int mindis=nodenum-1;
for(int v=0;v<nodenum;v++)
if(maze[u][v]-flow[u][v]&&mindis>dis[v]){
cur[u]=v;
mindis=dis[v];
}
if((--gap[dis[u]])==0) break;
gap[dis[u]=mindis+1]++;
u=pre[u];

}
return maxflow;
}

int main(){
int m,n;
int i,j,k,a,b,c;
while(~scanf("%d%d",&n,&m)){
memset(maze,0,sizeof(maze));
for(i=0;i<n;i++){
scanf("%d%d%d",&a,&b,&c);
maze[a][b]+=c;
}
int orz=sap(1,m,m+1);
printf("%d\n",orz);
}
}

邻接矩阵版：
#include<stdio.h>
#include<string.h>

const int MAXN = 100010 ; //点数最大值
const int MAXM = 400010 ; //边数最大值
const int INF = 0x3f3f3f3f;

struct Edge{
int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init(){
tol = 0;
memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w,int rw=0){
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].next = head[u];
edge[tol].flow = 0;
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].next = head[v];
edge[tol].flow = 0;
head[v] = tol++;
}

int sap(int start,int end,int N){
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while(dep[start] < N){
if(u==end){
int Min = INF;
for(int i=pre[u];i!= -1; i=pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;

for(int i=pre[u];i!=-1;i=pre[edge[i^1].to]){
edge[i].flow += Min;
edge[i^1].flow -=Min;
}
u=start;
ans +=Min;
continue;
}
bool flag = false;
int v;
for(int i= cur[u];i!=-1;i=edge[i].next){
v=edge[i].to;
if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u]){
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag){
u=v;
continue;
}
int Min = N;
for(int i=head[u];i!= -1;i=edge[i].next)
if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min){
Min=dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u] = Min +1;
gap[dep[u]]++;
if(u!=start) u = edge[pre[u]^1].to;
}
return ans;
}

int main(){
int m,n;
int i,j,k,a,b,c;
while(~scanf("%d%d",&n,&m)){
init();
for(i=0;i<n;i++){
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);
}
int orz=sap(1,m,m);
printf("%d\n",orz);
}
}

ORZ。。大神是自己写板。。我只能用。。差距好大。。。