HDU 4406

来源：互联网发布：矩阵潜袭有手机编辑：程序博客网时间：2024/06/09 16:26

费用流

建图还是比较简单的。。。（参见下面代码）

#include<cstdio>#include<cstring>#include<queue>#define N 100#define inf 100000000using namespace std;int cre[N],tot_cre;int sco[N];int head[N],cnt,dis[N],pre[N],vis[N],ans[N];int n;struct Edge{    int u,v,c,w,next;}edge[80010];void addedge(int u,int v,int cap,int cost){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].c=cap;    edge[cnt].w=cost;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].u=v;    edge[cnt].v=u;    edge[cnt].c=0;    edge[cnt].w=-cost;    edge[cnt].next=head[v];    head[v]=cnt++;}bool spfa(){    int i,j;    queue<int>q;    q.push(0);    for(i=1;i<=n;i++)        dis[i]=-inf,vis[i]=0;    dis[i]=0,vis[0]=1;    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(edge[i].c && dis[v] < dis[u]+edge[i].w){                dis[v]=dis[u]+edge[i].w;                pre[v]=i;                if(!vis[v]){                    vis[v]=1;                    q.push(v);                }            }        }    }    if(dis[n]==-inf)return 0;    return 1;}void end(){    int u,p,cap=-1;    for(u=n;u;u=edge[p].u){        p=pre[u];        if(cap==-1 || cap>edge[p].c)            cap=edge[p].c;    }    for(u=n;u;u=edge[p].u){        p=pre[u];        edge[p].c-=cap;        edge[p^1].c+=cap;    }}void init(){    memset(head,-1,sizeof(head));    memset(ans,0,sizeof(ans));    cnt=tot_cre=0;}int main(){    int i,j;    int v,e,k;    int pre,now,tem;    while(scanf("%d %d %d",&v,&k,&e)){        if(v==0 && k==0 && e==0)break;        init();        for(i=1;i<=e;i++){            scanf("%d",&cre[i]);            tot_cre+=cre[i];        }        n=v+e+1;        for(i=1;i<=v;i++)            addedge(0,i,k,0);        for(i=1;i<=e;i++){            scanf("%d",&sco[i]);            if(sco[i]<60){                pre=6400-3*(100-60)*(100-60);                addedge(i+v,n,60-sco[i],inf);                for(j=61;j<=100;j++){                    now=6400-3*(100-j)*(100-j);                    addedge(i+v,n,1,(now-pre)*cre[i]);                    pre=now;                }            }            else{                pre=6400-3*(100-sco[i])*(100-sco[i]);                for(j=sco[i]+1;j<=100;j++){                    now=6400-3*(100-j)*(100-j);                    addedge(i+v,n,1,(now-pre)*cre[i]);                    pre=now;                }            }        }        for(i=1;i<=v;i++)            for(j=1;j<=e;j++){                scanf("%d",&tem);                if(tem)                    addedge(i,v+j,k,0);            }        while(spfa())            end();        for(i=head[n];i!=-1;i=edge[i].next)            ans[edge[i].v-v]+=edge[i].c;        bool flag=1;        int sum=0;        for(i=1;i<=e;i++){            if(ans[i]+sco[i]<60){                flag=0;                break;            }            sum+=(6400-3*(100-ans[i]-sco[i])*(100-ans[i]-sco[i]))*cre[i];        }        if(!flag)printf("0.000000\n");        else printf("%.6lf\n",(double)sum/tot_cre/1600);    }}

如果边权用的是double类型的话一定记得加上eps

见下面代码（注意加上//////的代码）

#include<cstdio>#include<cstring>#include<queue>#include<cmath>#define N 100#define inf 100000000using namespace std;int cre[N],tot_cre;int sco[N];int head[N],cnt,pre[N],vis[N],ans[N];double dis[N];int n;struct Edge{    int u,v,c,next;    double w;}edge[80010];void addedge(int u,int v,int cap,double cost){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].c=cap;    edge[cnt].w=cost;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].u=v;    edge[cnt].v=u;    edge[cnt].c=0;    edge[cnt].w=-cost;    edge[cnt].next=head[v];    head[v]=cnt++;}bool spfa(){    int i,j;    queue<int>q;    q.push(0);    for(i=1;i<=n;i++)        dis[i]=-inf,vis[i]=0;    dis[i]=0,vis[0]=1;    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(edge[i].c && dis[v]-dis[u]-edge[i].w<-1e-9){       /////                dis[v]=dis[u]+edge[i].w;                pre[v]=i;                if(!vis[v]){                    vis[v]=1;                    q.push(v);                }            }        }    }    if(fabs(dis[n]+inf)<1e-9)return 0;      /////    return 1;}void end(){    int u,p,cap=-1;    for(u=n;u;u=edge[p].u){        p=pre[u];        if(cap==-1 || cap>edge[p].c)            cap=edge[p].c;    }    for(u=n;u;u=edge[p].u){        p=pre[u];        edge[p].c-=cap;        edge[p^1].c+=cap;    }}void init(){    memset(head,-1,sizeof(head));    memset(ans,0,sizeof(ans));    cnt=tot_cre=0;}int main(){    int i,j;    int v,e,k;    int pre,now,tem;    while(scanf("%d %d %d",&v,&k,&e)){        if(v==0 && k==0 && e==0)break;        init();        for(i=1;i<=e;i++){            scanf("%d",&cre[i]);            tot_cre+=cre[i];        }        n=v+e+1;        for(i=1;i<=v;i++)            addedge(0,i,k,0);        for(i=1;i<=e;i++){            scanf("%d",&sco[i]);            if(sco[i]<60){                pre=6400-3*(100-60)*(100-60);                addedge(i+v,n,60-sco[i],inf);                for(j=61;j<=100;j++){                    now=6400-3*(100-j)*(100-j);                    addedge(i+v,n,1,((double)now-pre)*cre[i]/(tot_cre*1600));                    pre=now;                }            }            else{                pre=6400-3*(100-sco[i])*(100-sco[i]);                for(j=sco[i]+1;j<=100;j++){                    now=6400-3*(100-j)*(100-j);                    addedge(i+v,n,1,((double)now-pre)*cre[i]/(tot_cre*1600));                    pre=now;                }            }        }        for(i=1;i<=v;i++)            for(j=1;j<=e;j++){                scanf("%d",&tem);                if(tem)                    addedge(i,v+j,k,0);            }        while(spfa())            end();        for(i=head[n];i!=-1;i=edge[i].next)            ans[edge[i].v-v]+=edge[i].c;        bool flag=1;        double sum=0;        for(i=1;i<=e;i++){            if(ans[i]+sco[i]<60){                flag=0;                break;            }            sum+=((double)6400-(double)3*(100-ans[i]-sco[i])*(100-ans[i]-sco[i]))*cre[i]/(tot_cre*1600);        }        if(!flag)printf("0.000000\n");        else printf("%.6lf\n",sum);    }}