Sicily 1696 Flows in Grid(最大流)

//由于给定的点最高可达100*100=10000//因此不可能建立一个10000*10000的图来描述流量，必定MLE//但考虑到是格子形状，每一个最多只与周边4个点相连//可以通过这个性质进行图的压缩，并建立一个方向和点的转化关系//再套用最大流模板便可通过#include<iostream>#include<queue>#include<cstring>using namespace std;const int MAX = 102,INF = 2000000000;int flow[MAX*MAX][4],cap[MAX*MAX][4],rflow[MAX*MAX];int pre[MAX*MAX];int c,maxflow;int st,ed,t,n,m;inline int Dir(int u,int v)//通过两点的关系转化返回方向{if(v - u == 1)return 0;if(u - v == 1)return 2;if(v - m == u)return 1;if(v + m == u)return 3;}inline int trans(int u,int v)//通过点和方向的关系返回坐标{if(v == 0)return u + 1;if(v == 1)return u + m;if(v == 2)return u - 1;if(v == 3)return u - m;}int main(){//freopen("in.txt","r",stdin);scanf("%d",&t);while(t--){memset(flow,0,sizeof(flow));memset(cap,0,sizeof(cap));scanf("%d%d",&n,&m);st = 1;ed = n*m;for(int i = 1;i <= n;++i){for(int j = 1;j <= m - 1;++j){scanf("%d",&c);cap[j + (i-1)*m][0] = cap[j + (i-1)*m + 1][2] = c;}}for(int i = 1;i <= n - 1;++i){for(int j = 1;j <= m;++j){scanf("%d",&c);cap[(i-1)*m + j][1] = cap[i * m + j][3] = c;}}maxflow = 0;for(;;)//最大流模板{queue<int> q;memset(rflow,0,sizeof(rflow));rflow[st] = INF;q.push(st);while(!q.empty()){int u = q.front();q.pop();for(int v = 0;v < 4;++v){if(!rflow[trans(u,v)] && cap[u][v] > flow[u][v]){q.push(trans(u,v));pre[trans(u,v)] = u;rflow[trans(u,v)] = min(rflow[u],cap[u][v] - flow[u][v]);}}}if(rflow[ed] == 0)break;for(int u = ed; u != st;u = pre[u]){flow[pre[u]][Dir(pre[u],u)] += rflow[ed];flow[u][Dir(u,pre[u])] -= rflow[ed];}maxflow += rflow[ed];}printf("%d/n",maxflow);}return 0;}