A - Drainage Ditches
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.
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
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10
Sample Output
50
题意： 给一个有向有环图，给出每条边到容量上限，无下限，源点是1，汇点是n，求最大流。赤裸裸点网络流，我用的ISAP算法。第一次过点网络流^_^
思路： ISAP模板过。白书上没给ISAP的BFS。。搞了好久才知道怎么改。。
代码
#include <vector>
#include <cstdio>
#include <cstring>
#include <queue>
#define FOR(i,n) for(i=1;i<=(n);i++)
using namespace std;
const int INF = 2e9+;
const int N = ; struct Edge{
int from,to,cap,flow;
}; struct ISAP{
int n,m,s,t;
int p[N],num[N];
vector<Edge> edges;
vector<int> G[N];
bool vis[N];
int d[N],cur[N];
void init(int _n,int _m)
{
n=_n; m=_m;
int i;
edges.clear();
FOR(i,n)
{
G[i].clear();
d[i]=INF;
}
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,});
edges.push_back((Edge){to,from,,});
m = edges.size();
G[from].push_back(m-);
G[to].push_back(m-);
}
bool BFS()
{
memset(vis,,sizeof(vis));
queue<int> Q;
Q.push(t);
d[t]=;
vis[t]=;
while(!Q.empty())
{
int x = Q.front(); Q.pop();
for(unsigned i=;i<G[x].size();i++)
{
Edge& e = edges[G[x][i]^];
if(!vis[e.from] && e.cap>e.flow)
{
vis[e.from]=;
d[e.from] = d[x]+;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment()
{
int x=t, a=INF;
while(x!=s)
{
Edge& e = edges[p[x]];
a = min(a,e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x!=s)
{
edges[p[x]].flow+=a;
edges[p[x]^].flow-=a;
x=edges[p[x]].from;
}
return a;
}
int Maxflow(int _s,int _t)
{
s=_s; t=_t;
int flow = , i;
BFS();
// FOR(i,n) printf("%d ",d[i]); puts("");
if(d[s]>=n) return ;
memset(num,,sizeof(num));
memset(p,,sizeof(p));
FOR(i,n) if(d[i]<INF) num[d[i]]++;
int x=s;
memset(cur,,sizeof(cur));
while(d[s]<n)
{
if(x==t)
{
flow+=Augment();
x=s;
}
int ok=;
for(unsigned i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow && d[x]==d[e.to]+)
{
ok=;
p[e.to]=G[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok)
{
int m=n-;
for(unsigned i=;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==) break;
num[d[x]=m+]++;
cur[x]=;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
}; ISAP isap; int main()
{
freopen("in","r",stdin);
int n,m,u,v,c;
while(scanf("%d%d",&m,&n)!=EOF)
{
isap.init(n,m);
while(m--)
{
scanf("%d%d%d",&u,&v,&c);
isap.AddEdge(u,v,c);
//isap.AddEdge(v,u,c);
}
printf("%d\n",isap.Maxflow(,n));
}
return ;
}
ISAP 模板
注意用宏定义的FOR来做点的初始化，有些题目点所从0开始编号有些所从1开始，所以需要用一个宏定义
struct Edge{
int from,to,cap,flow;
};
struct ISAP{
int n,m,s,t;
int p[N],num[N];
vector<Edge> edges;
vector<int> G[N];
bool vis[N];
int d[N],cur[N];
void init(int _n,int _m)
{
n=_n; m=_m;
int i;
edges.clear();
FOR(i,n)
{
G[i].clear();
d[i]=INF;
}
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,});
edges.push_back((Edge){to,from,,});
m = edges.size();
G[from].push_back(m-);
G[to].push_back(m-);
}
bool BFS()
{
memset(vis,,sizeof(vis));
queue<int> Q;
Q.push(t);
d[t]=;
vis[t]=;
while(!Q.empty())
{
int x = Q.front(); Q.pop();
for(unsigned i=;i<G[x].size();i++)
{
Edge& e = edges[G[x][i]^];
if(!vis[e.from] && e.cap>e.flow)
{
vis[e.from]=;
d[e.from] = d[x]+;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment()
{
int x=t, a=INF;
while(x!=s)
{
Edge& e = edges[p[x]];
a = min(a,e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x!=s)
{
edges[p[x]].flow+=a;
edges[p[x]^].flow-=a;
x=edges[p[x]].from;
}
return a;
}
int Maxflow(int _s,int _t)
{
s=_s; t=_t;
int flow = , i;
BFS();
if(d[s]>=n) return ;
memset(num,,sizeof(num));
memset(p,,sizeof(p));
FOR(i,n) num[d[i]]++;
int x=s;
memset(cur,,sizeof(cur));
while(d[s]<n)
{
if(x==t)
{
flow+=Augment();
x=s;
}
int ok=;
for(unsigned i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow && d[x]==d[e.to]+)
{
ok=;
p[e.to]=G[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok)
{
int m=n-;
for(unsigned i=;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==) break;
num[d[x]=m+]++;
cur[x]=;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
};