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POJ 1273
Drainage Ditches
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
Source
USACO 93
代码：
Dinic
邻接矩阵无优化#include<cstdio>#include<cstring>#define N 220int map[N][N],h[N],s,t,e,c,data[N],n,m;int min(int x,int y){return x<y?x:y;}bool bfs(){memset(h,-1,sizeof(h));//从源点编号 data[1]=s;int op=0,cl=1;h[s]=0;bool flag=false;while (op<cl){++op;int u=data[op];for (int i=1;i<=n;++i)if (map[u][i]>0) if (h[i]==-1){h[i]=h[u]+1;data[++cl]=i;if (i==t) flag=true;}}return flag;}int dfs(int u,int s){//初值if (u==n) return s;int tmp=s;for (int i=1;i<=n;++i){if ((map[u][i]>0)&&(h[u]+1==h[i])) {int x=dfs(i,min(s,map[u][i]));map[u][i]-=x;map[i][u]+=x;s-=x;if (s==0) break;}} return tmp-s;}int main(){freopen("poj1273.in","r",stdin);freopen("poj1273.out","w",stdout);int maxlong=0x7fffffff;while (scanf("%d%d",&m,&n)>0){memset(map,0,sizeof(map));for (int i=1;i<=m;++i){scanf("%d%d%d",&s,&e,&c);map[s][e]+=c;}//源点1 汇点1 s=1;t=n;int ans=0,x=0;while (bfs()){x=dfs(s,maxlong);ans+=x;}printf("%d\n",ans);}return 0;}
邻接矩阵当前弧优化
#include<cstdio>#include<cstring>#include<iostream>//#include<algorithm>using namespace std;int n,m,map[220][220],h[220],data[220],cur[220];bool bfs(){memset(h,-1,sizeof(h));data[1]=1;h[1]=0;int op=0,cl=1;while (op<cl){++op;int u=data[op];for (int v=1;v<=n;++v){if (map[u][v]!=0&&h[v]==-1){data[++cl]=v;h[v]=h[u]+1;if (v==n) return true;}}}return false;}//int min(int x,int y){// return x<y?x:y;//}int dfs(int u,int s){if (u==n) return s;int ss=s;int y=1;for (int v=cur[u];v<=n;++v){if (map[u][v]>0&&h[v]==h[u]+1){int x=dfs(v,min(map[u][v],s));s-=x;map[u][v]-=x;map[v][u]+=x;y=v;if (s==0) {cur[u]=v;return ss;}}}cur[u]=y;return ss-s;}int main(){freopen("poj1743.in","r",stdin);freopen("poj1743.out","w",stdout);while (scanf("%d%d",&m,&n)>0){memset(map,0,sizeof(map));for (int i=1;i<=m;++i){int x,y,z;scanf("%d%d%d",&x,&y,&z);map[x][y]+=z;}int ans=0;while (bfs()){for (int i=1;i<=n;++i) cur[i]=1;int xx=dfs(1,0x7fffffff);ans+=xx;}printf("%d\n",ans);}return 0;}
邻接表无优化
#include<cstdio>#include<cstring>#include<iostream>//#include<algorithm>using namespace std;struct node{int x,y,z,next;};node data1[440];int n,m,hh[220],h[220],data[220];bool bfs(){memset(hh,-1,sizeof(hh));data[1]=1;hh[1]=0;int op=0,cl=1;while (op<cl){++op;int u=data[op];for (int i=h[u];i!=-1;i=data1[i].next){int v=data1[i].y;if (data1[i].z>0&&hh[v]==-1){data[++cl]=v;hh[v]=hh[u]+1;if (v==n) return true;} }}return false;}//int min(int x,int y){// return x<y?x:y;//}int dfs(int u,int s){if (u==n) return s;int ss=s;for (int i=h[u];i!=-1;i=data1[i].next){int v=data1[i].y;if (data1[i].z>0&&hh[v]==hh[u]+1){int x=dfs(v,min(data1[i].z,s));s-=x;data1[i].z-=x;data1[i^1].z+=x;if (s==0) return ss;}}return ss-s;}int main(){// freopen("poj1273.in","r",stdin);// freopen("poj1273.out","w",stdout);while (scanf("%d%d",&m,&n)>0){//memset(map,0,sizeof(map));memset(h,-1,sizeof(h));int num=-1;for (int i=1;i<=m;++i){int x,y,z;scanf("%d%d%d",&x,&y,&z);data1[++num].x=x;data1[num].y=y;data1[num].z=z;data1[num].next=h[x];h[x]=num;data1[++num].x=y;data1[num].y=x;data1[num].next=h[y];h[y]=num;data1[num].z=0;//map[x][y]+=z;}int ans=0;while (bfs()){int xx=dfs(1,0x7fffffff);//printf("%d\n",xx);ans+=xx;}printf("%d\n",ans);}return 0;}
邻接表当前弧优化
#include<cstdio>#include<cstring>struct node{int x,y,z,next;};node data[440];int cur[220],f[220],m,n,h[220],st[220];int min(int x,int y){return x<y?x:y;} bool bfs(){memset(f,-1,sizeof(f));st[1]=1;f[1]=0;int op=0,cl=1;while(op<cl){op++;int u=st[op];for (int i=h[u];i!=-1;i=data[i].next){int v=data[i].y;if (data[i].z>0&&f[v]==-1){f[v]=f[u]+1;st[++cl]=v;if (v==n)return true;}}}return false;}int dfs(int u,int s){if (u==n) return s;int y=h[u];int tmp=s;for (int i=cur[u];i!=-1;i=data[i].next){int v=data[i].y;if (data[i].z>0&&f[v]==f[u]+1){int x=dfs(v,min(s,data[i].z));s-=x;y=i;data[i].z-=x;data[i^1].z+=x;if (s==0){cur[u]=i;return tmp;}}}cur[u]=y; return tmp-s;}int main(){freopen("poj1273.in","r",stdin);freopen("poj1273.out","w",stdout);while (scanf("%d%d",&m,&n)>0){int num=-1;memset(h,-1,sizeof(h));for (int i=1;i<=m;++i){int s,e,c;scanf("%d%d%d",&s,&e,&c);data[++num].x=s;data[num].y=e;data[num].z=c;data[num].next=h[s];h[s]=num;data[++num].x=e;data[num].y=s;data[num].next=h[e];h[e]=num;}int ans=0;for (int i=1;i<=n;++i){cur[i]=h[i];}while(bfs()){int x=dfs(1,0x7fffffff);ans+=x;for (int i=1;i<=n;++i){cur[i]=h[i];} }printf("%d\n",ans);}return 0;}
sap
想法
Int sap
到u是t的时候返回水量
先备份
给所有子边灌水
1 需要有边，并且只差1
2 求可以给下一个的送水量
如果该点无法送水 找所有有关联的h的最小值(不能饱和,也必须有边)
否则返回s1-s
