HDU 3572 Task Schedule (DINIC 邻接表实现)

来源：互联网发布：政法工作大数据对策编辑：程序博客网时间：2024/06/09 18:42
YY了好久才弄出来。。。。每次问题都被喷。。。好伤心。。。快哭了
#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;#define N 1005#define INF 0x7ffffffint n,m,k;int level[N];struct node{    int to,next,cost;}edge[N*N];int head[N];int t_head[N];int start,end;void init(){    k=0;    memset(head,-1,sizeof(head));}int bfs(int s,int t)//对顶点进行标号、找出层次图{    memset(level,0,sizeof(level));    queue<int>q;    q.push(s);    level[s]=1;    while(!q.empty())    {        int now=q.front();        q.pop();        for(int i=head[now];i!=-1;i=edge[i].next)        {            int y=edge[i].to;            if(!level[y]&&edge[i].cost>0)            {                level[y]=level[now]+1;                q.push(y);            }        }    }    return level[t]!=0;//汇点是否在层次图中}int dfs(int s,int cp)//在层次图中寻找增广路进行增广{    int flow=0,temp;    int t;    if(s==end||cp==0)return cp;    for(;t_head[s]+1;t_head[s]=edge[t_head[s]].next)    {        int y=edge[t_head[s]].to;        if(level[s]+1==level[y])        {            temp=dfs(y,min(cp,edge[t_head[s]].cost));            if(temp>0)            {                edge[t_head[s]].cost-=temp;                edge[t_head[s]^1].cost+=temp;                flow+=temp;                cp-=temp;                if(cp==0)break;            }        }    }    return flow;}int dinic(){    int ans=0,flow=0;    while(bfs(start,end))//汇点不在层次图中、算法终止    {        for(int i=0;i<=end;i++)            t_head[i]=head[i];        ans+=dfs(start,INF);    }    return ans;}void add(int x,int y,int val){    edge[k].to=y;    edge[k].cost=val;    edge[k].next=head[x];    head[x]=k++;    edge[k].to=x;    edge[k].cost=0;    edge[k].next=head[y];    head[y]=k++;}int main(){    int t;    int cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        init();        int sum=0;        start=0,end=500+n+1;        for(int i=1;i<=n;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            sum+=x;            add(start,i,x);            for(int j=y;j<=z;j++)                add(i,j+n,1);        }        for(int i=1;i<=500;i++)            add(i+n,end,m);        int ans=dinic();        printf("Case %d: ",cas++);        if(ans!=sum)printf("No\n\n");        else printf("Yes\n\n");    }    return 0;}
0 0