HDU 5045 Contest (cost stream), hdu5045

HDU 5045 Contest (cost stream), hdu5045

Address: HDU 5045

Finally, I used network stream A in the competition... I used so many network streams once .. Although the question is very simple, I still want to remember it...

My idea for this question is to find m/n charge streams, and calculate the maximum charge flow for the same round for every n of them. Creating a graph is simple. It is the simplest Binary Graph Model.

The Code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64using namespace std;double mp[20][2000], d[200], cost;const int INF=0x3f3f3f3f;int head[200], source, sink, cnt, flow;int f[200], cur[200], vis[200];struct node{    int u, v, cap, next;    double cost;} edge[100000];void add(int u, int v, int cap, double 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(){    int i;    for(i=0; i<200; i++)        d[i]=INF;    memset(vis,0,sizeof(vis));    cur[source]=-1;    f[source]=INF;    d[source]=0;    queue<int>q;    q.push(source);    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;                f[v]=min(f[u],edge[i].cap);                cur[v]=i;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    if(d[sink]==INF) return 0;    flow+=f[sink];    cost-=f[sink]*d[sink];    for(i=cur[sink]; i!=-1; i=cur[edge[i^1].v])    {        edge[i].cap-=f[sink];        edge[i^1].cap+=f[sink];    }    return 1;}void mcmf(){    cost=flow=0;    while(spfa()) ;}int main(){    int t, num=0, i, j, k, n, m;    double sum;    scanf("%d",&t);    while(t--)    {        sum=0;        num++;        scanf("%d%d",&n,&m);        for(i=1; i<=n; i++)        {            for(j=1; j<=m; j++)            {                scanf("%lf",&mp[i][j]);            }        }        k=1;        sum=0;        while(k<=m)        {            source=0;            sink=2*n+1;            memset(head,-1,sizeof(head));            cnt=0;            for(i=k; i<k+n&&i<=m; i++)            {                add(i-k+1+n,sink,1,0);                for(j=1; j<=n; j++)                {                    add(j,i-k+1+n,1,-mp[j][i]);                }            }            for(i=1; i<=n; i++)                add(source,i,1,0);            mcmf();            sum+=cost;            //printf("%.2lf    %d\n",sum, k);            k+=n;        }        printf("Case #%d: %.5lf\n",num,sum);    }    return 0;}

Calculate the distance between two points. Question: http: // acmhdueducn/diy/contest_showproblemphp? Pid = 1002 & cid = 12905 & hide = 0

If you submit c ++, submit it to gcc or c.

Acm problem http: // acmhdueducn/contests/contest_showproblemphp? Pid = 1009 & cid = 389 hangdian, what's the idea?

It seems that you can create a vertical line of three edges first. After obtaining the center point, each edge and the center point form three triangles. The three triangles correspond to the three corners of the center vertex to obtain the maximum approximate number, divide by 360 by the maximum common divisor, which is the number of positive polygon edges of the Minimum side. We hope you can achieve this.