GPA

GPA(Grade-Point Average) is one way to measure students’ academic performance in PKU. Each course has an integer credit, ranges from 1 to 99. For each course, you will get a score at the end of the semester, which is an integer ranges from 0 to 100. Then you can calculate the Grade-Point of this course with the following formula. (Your score is x and your Grade-Point is p, using real arithmetic)

Then you can get the GPA with the following formula (the Grade-Point of course i is p_i, and the credit of course i is w_i).

Now it is not far from the final exam, if you do not review, you can only get a basic score in each course.

You have n days to review. There are K classes in each day. For each class, only one course can be reviewed. After the review, your score in this course will exactly increase by 1. You can get more increment by spending more classes in this course. But the score may not exceed 100.

For some reasons, not any course can be reviewed in any class. Each day you can only review some of the courses.

Now you want your GPA to be as high as possible, and at the same time, you do not want to fail in any course. Please calculate the highest GPA you can get. 

2 10 3
1 1 2
50 60 90
1 1 0
1 0 1
2 20 4
1 1 1 1
50 50 50 40
1 1 1 0
0 0 0 1
0 0 0

2.757813
0.000000

有m门课，n天，每天可以复习特定的几门课，每门课有学分，基础分，复习一次可以加分，绩点有计算公式，问你总绩点最多多少？

用最大费用流可以做。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int maxn=11000;
const int maxm=1100000;
const double inf=1e8;

struct edge{
    int u,v,next,f;
    double c;
    edge(int u0=0,int v0=0,int f0=0,double c0=0,int next0=0){
        u=u0,v=v0,f=f0,c=c0,next=next0;
    }
}e[maxm];

int head[maxn],path[maxn];
double dist[maxn];
bool visited[maxn];
int cnt,src,sink;

void init(){
    cnt=0;
    memset(head,-1,sizeof(head));
}

void adde(int u,int v,int f,double c){
    e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].c=c,e[cnt].next=head[u],head[u]=cnt++;
    e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].c=-c,e[cnt].next=head[v],head[v]=cnt++;
}

bool bfs(){
     for(int i=src;i<=sink;i++){
        dist[i]=-inf;
        path[i]=-1;
     }
     dist[src]=0;
     queue <int> q;
     q.push(src);
     visited[src]=true;
     while(!q.empty()){
         int s=q.front();
         q.pop();
         for(int i=head[s];i!=-1;i=e[i].next){
              int d=e[i].v;
              if(e[i].f>0 && dist[s]+e[i].c>dist[d]){
                 dist[d]=dist[s]+e[i].c;
                 path[d]=i;
                 if(!visited[d]){
                    visited[d]=true;
                    q.push(d);
                 }
              }
         }
         visited[s]=false;
     }
     return path[sink]>=0;
}

double getMaxCost(){
    //int maxflow=0;
    double ret=0;
    while(bfs()){
        int delta=inf;
        for(int i=sink;i!=src;i=e[path[i]].u){
            if( e[path[i]].f<delta ) delta=e[path[i]].f;
        }
        for(int i=sink;i!=src;i=e[path[i]].u){
            e[path[i]].f-=delta;
            e[path[i]^1].f+=delta;
        }
        ret+=dist[sink]*delta;
        //maxflow+=delta;
    }
    //cout<<"maxflow:"<<maxflow<<endl;
    return ret;
}

int nd,nf,nc;
int fen[50],base[50],flag[50][50];
const double eps=1e-6;

void input(){
    for(int i=0;i<nc;i++) scanf("%d",&fen[i]);
    for(int i=0;i<nc;i++) scanf("%d",&base[i]);
    for(int i=0;i<nd;i++){
        for(int j=0;j<nc;j++){
            scanf("%d",&flag[i][j]);
        }
    }
}

double getf(double x){
    return 4-3*(100-x)*(100-x)/1600.0;
}

void build(){
    init();
    src=0,sink=nc+nd+1;
    for(int i=0;i<nc;i++){
        if(base[i]<60){
            adde(src,i+1,60-base[i],inf);
            for(int t=60;t<100;t++){
                adde(src,i+1,1, ( getf(t+1)-getf(t) )*fen[i] );
            }
        }else{
            for(int t=base[i];t<100;t++){
                adde(src,i+1,1, ( getf(t+1)-getf(t) )*fen[i] );
            }
        }
    }

    for(int i=0;i<nd;i++){
        for(int j=0;j<nc;j++){
            if(flag[i][j]){
                adde(j+1,nc+i+1,nf,0);
            }
        }
    }
    for(int i=0;i<nd;i++){
        adde(i+1+nc,sink,nf,0);
    }
}

void solve(){
    build();
    double ans=getMaxCost();
    for(int i=head[src];i!=-1;i=e[i].next){
        if(e[i].f>0 && e[i].c>inf-eps){
            printf("0.000000\n");
            return;
        }
    }
    int sum=0;
    for(int i=0;i<nc;i++) sum+=fen[i];
    for(int i=0;i<nc;i++){
        if(base[i]<60){
            ans-=inf*(60-base[i]);
            ans+=( getf(60)-getf(base[i]) )*fen[i];
        }
        ans+=getf(base[i])*fen[i];
    }
    printf("%.6lf\n",ans/sum);
}

int main(){
    while(scanf("%d%d%d",&nd,&nf,&nc)!=EOF && (nd||nf||nc) ){
        input();
        solve();
    }
    return 0;
}

HDU 4406 GPA（网络流-最大费用流）,布布扣,bubuko.com

HDU 4406 GPA（网络流-最大费用流）

原文：http://blog.csdn.net/a1061747415/article/details/38497927