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Course Selection Problem Code: RIN
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Read problems statements in Mandarin Chineseand Russian.
Rin is attending a university.
She has M semesters to finish her program, and that program has N required courses. Each course must be taken in exactly one of the semesters.
Some courses have prerequisites: for each i from 1 to K, she must take course A[i]before course B[i].
The same course may be taught by different professors in different semesters, and how well she does will depend on which professor teaches her. Additionally, some courses may not be taught every semester.
We are given an array X representing this information. For each course i and each semester j, X[i][j] = -1 if course i is not taught in semester j. Otherwise, X[i][j] will be an integer between 0 and 100: the expected score Rin will get if she takes course i in semester j.
Rin may take any number of courses per semester, including none, as long as they are taught that semester and she has already taken any required prerequisite courses.
Help Rin to find the maximal average score she can get over her whole program.
Input
The first line contain 3 integers: N, M and K.
This is followed by N lines, each containing M integers. The jth integer on the ith line represents the value of X[i][j].
This is followed by K lines, each containing two integers: A[i] and B[i].
Output
Output one real number: the maximal average score, rounded to 2 digits after the decimal point.
Constraints
- 1 ≤ M, N ≤ 100
- 0 ≤ K ≤ 100
- -1 ≤ X[i][j] ≤ 100
- 1 ≤ A[i], B[i] ≤ N
- For each i, A[i] ≠ B[i].
- For different i and j, (A[i], B[i]) ≠ (A[j], B[j]).
- We guarantee there exists a way to take these N courses in M semesters.
Subtasks
Subtask 1: (20 Points) A course can have at most 1 pre request course.
Subtask 2: (80 Points) Refer to constraints above
Example
Input 1: 3 2 2 70 100 100 80 100 90 1 2 1 3 Output 1: 80.00 Input 2: 4 5 4 20 -1 100 -1 -1 100 30 -1 -1 -1 100 -1 30 20 40 100 30 40 50 20 1 2 1 3 2 4 3 4 Output 2: 32.50
Explanation
Example case 1
The only way she can finish these 3 courses is: take course 1 in the first semester, then take courses 2 and 3 in the second semester. The average score is (70 + 80 + 90) / 3 = 80.00.
Example case 2
The optimal solution is: take course 1 in semester 1, course 2 in semester 2, course 3 in semester 3 and course 4 in semester 4.
n个课程要在m个学期里完成，
有k个依赖关系a，b ，表示a课程必须在b课程之前
每个课程在每个学期有相应得分
-1 表示本学期不开设此课程
问 最大的课程得分平均值
最大得分转化为最小扣分
对下面这张图跑最小割，ans=n*100-最小割
解释：Step 3 ：防止所有边都割在了源点向第一学期
Step 4 : inf边一定不会被割，a是b的前置课程，如果b的边割在了a之前，图可以通过inf边联通，不符合最小割
求助大佬，为什么以下建图方式不对
#include<queue>
#include<cstdio>
#include<algorithm>
#define inf 0x7fffffff
#define N 10101
#define M 50000
using namespace std;
queue<int>q;
int n,m,k,src,decc,tmp;
int front[N],to[M],nxt[M],tot=1;
int lev[N],cur[N],cap[M];
int score[101][101];
bool f[101];
//#define turn((i),(j)) ((i)-1)*(m+1)+(j)
int turn(int i,int j)
{
return (i-1)*m+j;
}
void add(int u,int v,int w)
{
to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; cap[tot]=w;
to[++tot]=u; nxt[tot]=front[v]; front[v]=tot;
}
bool bfs()
{
while(!q.empty()) q.pop();
for(int i=src;i<=decc;i++) cur[i]=front[i],lev[i]=-1;
lev[src]=0; q.push(src);
int now;
while(!q.empty())
{
now=q.front(); q.pop();
for(int i=front[now];i;i=nxt[i])
{
if(lev[to[i]]==-1&&cap[i]>0)
{
lev[to[i]]=lev[now]+1;
if(to[i]==decc) return true;
q.push(to[i]);
}
}
}
return false;
}
int dinic(int now,int flow)
{
if(now==decc) return flow;
int rest=0,delta;
for(int &i=cur[now];i;i=nxt[i])
{
if(lev[to[i]]>lev[now]&&cap[i]>0)
{
delta=dinic(to[i],min(flow-rest,cap[i]));
if(delta)
{
cap[i]-=delta; cap[i^1]+=delta;
rest+=delta; if(rest==flow) break;
}
}
}
if(rest!=flow) lev[now]=-1;
return rest;
}
int main()
{
scanf("%d%d%d",&n,&m,&k);
int x; decc=n*m+1;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&score[i][j]);
for(int i=1;i<=n;i++) add(turn(i,m),decc,inf);
int a,b;
while(k--)
{
scanf("%d%d",&a,&b);
for(int i=1;i<m;i++) add(turn(a,i),turn(b,i+1),inf);
f[b]=true;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(j==1)
{
if(f[i]) add(src,turn(i,1),inf);
else add(src,turn(i,1),score[i][j]==-1 ? inf : 100-score[i][j]);
}
else add(turn(i,j-1),turn(i,j),score[i][j]==-1 ? inf : 100-score[i][j]);
while(bfs())
tmp+=dinic(src,inf);
tmp=n*100-tmp;
printf("%.2lf",double(tmp)/n);
}
错误建图代码
#include<queue>
#include<cstdio>
#include<algorithm>
#define inf 0x7fffffff
#define N 10101
#define M 50000
using namespace std;
queue<int>q;
int n,m,k,src,decc,tmp;
int front[N],to[M],nxt[M],tot=1;
int lev[N],cur[N],cap[M];
//#define turn((i),(j)) ((i)-1)*(m+1)+(j)
int turn(int i,int j)
{
return (i-1)*(m+1)+j;
}
void add(int u,int v,int w)
{
to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; cap[tot]=w;
to[++tot]=u; nxt[tot]=front[v]; front[v]=tot;
}
bool bfs()
{
while(!q.empty()) q.pop();
for(int i=src;i<=decc;i++) cur[i]=front[i],lev[i]=-1;
lev[src]=0; q.push(src);
int now;
while(!q.empty())
{
now=q.front(); q.pop();
for(int i=front[now];i;i=nxt[i])
{
if(lev[to[i]]==-1&&cap[i]>0)
{
lev[to[i]]=lev[now]+1;
if(to[i]==decc) return true;
q.push(to[i]);
}
}
}
return false;
}
int dinic(int now,int flow)
{
if(now==decc) return flow;
int rest=0,delta;
for(int &i=cur[now];i;i=nxt[i])
{
if(lev[to[i]]>lev[now]&&cap[i]>0)
{
delta=dinic(to[i],min(flow-rest,cap[i]));
if(delta)
{
cap[i]-=delta; cap[i^1]+=delta;
rest+=delta; if(rest==flow) break;
}
}
}
if(rest!=flow) lev[now]=-1;
return rest;
}
int main()
{
scanf("%d%d%d",&n,&m,&k);
int x; decc=n*(m+1)+1;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
scanf("%d",&x);
if(x==-1) x=inf;
else x=100-x;
add(turn(i,j),turn(i,j+1),x);
}
for(int i=1;i<=n;i++)
add(0,turn(i,1),inf),
add(turn(i,m+1),decc,inf);
int a,b;
while(k--)
{
scanf("%d%d",&a,&b);
for(int i=1;i<=m;i++) add(turn(a,i),turn(b,i+1),inf);
}
while(bfs())
tmp+=dinic(src,inf);
tmp=n*100-tmp;
printf("%.2lf",double(tmp)/n);
}