HDU 2686 & HDU 3376 (cost stream of network Stream)

Address: HDU 2686 HDU 3376

These two questions are the same except the data size. The former is only 30*30, but the latter is 600*600 .. I thought that the former Code used the latter would time out and did not dare to pay for it, but I felt that I could only do this if I could use the cost stream, so I changed the array size and handed it over. No timeout ..

This is also about the short circuit back and forth. You can change the maximum charge to the opposite, and then convert it to the opposite.

The idea of creating a graph is to split points, and each point can only go through once. Then, connect each vertex to the edge on the right and bottom. Set the source point and sink point to two traffic types.

I don't know why to use g ++ to keep Wa, and switch to C ++ to AC...

The Code is as follows;

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;int head[800000], source, sink, cost, flow, cnt, mp[700][700];int cur[800000], d[800000], pre[800000], vis[800000];struct node{    int u, v, cap, next, cost;}edge[10000000];void add(int u, int v ,int cap, int cost){    edge[cnt].v=v;    edge[cnt].cap=cap;    edge[cnt].cost=cost;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].cap=0;    edge[cnt].cost=-cost;    edge[cnt].next=head[v];    head[v]=cnt++;}int spfa(){    memset(d,INF,sizeof(d));    memset(vis,0,sizeof(vis));    queue<int>q;    q.push(source);    cur[source]=-1;    d[source]=0;    int minflow=INF, i;    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=0;        for(i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            if(d[v]>d[u]+edge[i].cost&&edge[i].cap)            {                d[v]=d[u]+edge[i].cost;                minflow=min(minflow,edge[i].cap);                cur[v]=i;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    if(d[sink]==INF) return 0;    flow+=minflow;    cost-=minflow*d[sink];    for(i=cur[sink];i!=-1;i=cur[edge[i^1].v])    {        edge[i].cap-=minflow;        edge[i^1].cap+=minflow;    }}void mcmf(int n){    cost=0;    flow=0;    while(spfa()) ;    printf("%d\n",cost-mp[0][0]-mp[n-1][n-1]);}int main(){    int n, i, j, k;    while(scanf("%d",&n)!=EOF)    {        memset(head,-1,sizeof(head));        cnt=0;        for(i=0;i<n;i++)        {            for(j=0;j<n;j++)            {                scanf("%d",&mp[i][j]);            }        }        source=2*n*n;        sink=2*n*n+1;        add(source,0,2,0);        add(2*n*n-1,sink,2,0);        for(i=0;i<n;i++)        {            for(j=0;j<n;j++)            {                if((i==0&&j==0)||(i==n-1&&j==n-1))                {                    add(i*n+j,i*n+j+n*n,2,-mp[i][j]);                }                else                    add(i*n+j,i*n+j+n*n,1,-mp[i][j]);                if(i<n-1)                    add(i*n+j+n*n,(i+1)*n+j,1,0);                if(j<n-1)                    add(i*n+j+n*n,i*n+j+1,1,0);            }        }        mcmf(n);    }    return 0;}

HDU 2686 & HDU 3376 (cost stream of network Stream)