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  • 最大流任务调度+离散化——hdu2883

     思想就是把时间段离散化,然后用个点来表示一段时间

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include <algorithm>
    #include <math.h>
    #include <queue>
    using namespace std;
    const int N = 1005;
    const int INF = 999999999;
    struct Edge{
        int v,w,next;
    }edge[N*N];
    int head[N];
    int level[N];
    int tot;
    void init()
    {
        memset(head,-1,sizeof(head));
        tot=0;
    }
    void addEdge(int u,int v,int w,int &k)
    {
        edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++;
        edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++;
    }
    int BFS(int src,int des)
    {
        queue<int>q;
        memset(level,0,sizeof(level));
        level[src]=1;
        q.push(src);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            if(u==des) return 1;
            for(int k = head[u]; k!=-1; k=edge[k].next)
            {
                int v = edge[k].v;
                int w = edge[k].w;
                if(level[v]==0&&w!=0)
                {
                    level[v]=level[u]+1;
                    q.push(v);
                }
            }
        }
        return -1;
    }
    int dfs(int u,int des,int increaseRoad){
        if(u==des||increaseRoad==0) return increaseRoad;
        int ret=0;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v,w=edge[k].w;
            if(level[v]==level[u]+1&&w!=0){
                int MIN = min(increaseRoad-ret,w);
                w = dfs(v,des,MIN);
                if(w > 0)
                {
                    edge[k].w -=w;
                    edge[k^1].w+=w;
                    ret+=w;
                    if(ret==increaseRoad) return ret;
                }
                else level[v] = -1;
                if(increaseRoad==0) break;
            }
        }
        if(ret==0) level[u]=-1;
        return ret;
    }
    int Dinic(int src,int des)
    {
        int ans = 0;
        while(BFS(src,des)!=-1) ans+=dfs(src,des,INF);
        return ans;
    }
    int n,m;
    int s[N],num[N],e[N],t[N];
    int time[N];
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF){
            init();
            int k = 0;
            int sum = 0;
            for(int i=1;i<=n;i++){
                scanf("%d%d%d%d",&s[i],&num[i],&e[i],&t[i]);
                sum += num[i]*t[i];
                time[++k] = s[i];
                time[++k] = e[i];
            }
            sort(time+1,time+k+1);
            int cnt = 1;
            for(int i=2;i<=k;i++){
                if(time[i]==time[i-1]) continue;
                time[++cnt] = time[i];
            }
            int src = 0,des = n+cnt+1;
            for(int i=1;i<=n;i++){
                addEdge(src,i,num[i]*t[i],tot);
            }
            for(int i=1;i<=n;i++){
                for(int j=1;j<=cnt;j++){
                    if(s[i]<=time[j-1]&&time[j]<=e[i]){
                        addEdge(i,n+j,INF,tot);
                    }
                }
            }
            for(int i=1;i<=cnt;i++){
                addEdge(n+i,des,(time[i]-time[i-1])*m,tot);
            }
            if(Dinic(src,des)==sum) printf("Yes
    ");
            else printf("No
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zsben991126/p/10992786.html
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