卧薪尝胆,厚积薄发。
八省联考2018 劈配
Date: Thu Mar 14 17:20:48 CST 2019 In Category: NoCategory

Description:

$n$ 个人按从 $1$ 到 $n$ 排序,每个人有 $m$ 档志愿,每档志愿最多有 $c$ 个导师,每个导师有人数上限,依次选导师,问每个人最优被哪一档支援录取,以及他排名提高几名可以被第 $s_i$ 档志愿录取。
$1\leqslant n\leqslant 200$

Solution:

首先第一问很好做,直接动态加边 $dinic$ 即可,对于第二问,我们可以暴力记录每个人选完之后的残量网络,然后二分提高的名次,再新建一个图加入这个人的边判断即可。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cctype>
#include<cstring>
using namespace std;
int T,C;
#define MAXN 210
#define INF 0x3f3f3f3f
int n,m;
int b[MAXN];
int s[MAXN];
vector<int> v[MAXN][MAXN];
int res[MAXN];
struct DINIC
{
struct edge
{
int to,nxt,f;
}e[MAXN * 40];
int edgenum;
int lin[MAXN * 2];
int ans;
DINIC(){ans = 0;memset(lin,0,sizeof(lin));edgenum = 0;}
void add(int a,int b,int f)
{
e[edgenum] = (edge){b,lin[a],f};lin[a] = edgenum++;
e[edgenum] = (edge){a,lin[b],0};lin[b] = edgenum++;
return;
}
int ch[MAXN * 2];
int s,t;
bool BFS()
{
queue<int> q;q.push(s);
memset(ch,-1,sizeof(ch));ch[s] = 0;
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return (ch[t] != -1);
}
int flow(int k,int f)
{
if(k == t)return f;
int r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f)
{
int l = flow(e[i].to,min(f - r,e[i].f));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r == 0)ch[k] = -1;
return r;
}
int dinic()
{
int r;
while(BFS())while(r = flow(s,INF))ans += r;
return ans;
}
void init()
{
edgenum = 0;
memset(lin,-1,sizeof(lin));
s = t = 0;
return;
}
}g[MAXN],p;
bool check(int k,int d)
{
p = g[k - d - 1];
for(int i = 1;i <= s[k];++i)
{
for(vector<int>::iterator it = v[k][i].begin();it != v[k][i].end();++it)
{
p.add(k,*it + n,1);
}
}
return (p.dinic() > g[k - d - 1].ans);
}
void work()
{
scanf("%d%d",&n,&m);
g[0].init();
g[0].s = n + m + 1;g[0].t = n + m + 2;
for(int i = 1;i <= m;++i)scanf("%d",&b[i]);
for(int i = 1;i <= m;++i)g[0].add(i + n,g[0].t,b[i]);
for(int i = 1;i <= n;++i)g[0].add(g[0].s,i,1);
for(int i = 1;i <= n;++i)for(int j = 1;j <= m;++j)v[i][j].clear();
int x;
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
scanf("%d",&x);
if(x != 0)v[i][x].push_back(j);
}
}
for(int i = 1;i <= n;++i)scanf("%d",&s[i]);
for(int i = 1;i <= n;++i)
{
int j;
for(j = 1;j <= m;++j)
{
g[i] = g[i - 1];
for(vector<int>::iterator it = v[i][j].begin();it != v[i][j].end();++it)
{
g[i].add(i,*it + n,1);
}
if(g[i].dinic() > g[i - 1].ans)break;
}
res[i] = j;
}
for(int i = 1;i <= n;++i)printf("%d ",res[i]);cout << endl;
for(int i = 1;i <= n;++i)
{
int l = 0,r = i,mid;
while(l < r)
{
mid = ((l + r) >> 1);
if(check(i,mid))r = mid;
else l = mid + 1;
}
printf("%d ",l);
}
cout << endl;
return;
}
int main()
{
scanf("%d%d",&T,&C);
while(T--)work();
return 0;
}
In tag: 图论-dinic
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