【luogu 2050】【NOI2012】美食节（费用流）
作者：互联网
【题目大意】n种菜，每种有pi份，有m位厨师，第i位厨师做第j道菜的时间是cij，求将所有菜做完花费的最小时间
【题目分析】
【相对暴力】
直接进行拆点，类似修车
（其实就是一样的
但是问题在于如果直接拆的话点数过多，就会TLE
于是就需要想办法进行优化
（建议不熟的话先把暴力写会）
// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cstring>
#define ll int
using namespace std;
const int MAXM = 113234;
const int INF = 2e9 + 7;
typedef pair<int, int> P;
struct note
{
int to;
int nt;
int rev;
ll cost;
ll cal;
};
int tn, tm, tk, sum;
int p[MAXM], a[200][200];
void read()
{
scanf("%d%d", &tn, &tm);
for (int i = 1; i <= tn; i++)
scanf("%d", &p[i]), sum += p[i];
for (int i = 1; i <= tn; i++)
for (int j = 1; j <= tm; j++)
scanf("%d", &a[i][j]);
}
struct edge
{
note arr[7000000];
int st[MAXM];
ll dis[MAXM];
ll h[MAXM];
int cur[MAXM];
int depth[MAXM];
int pre[MAXM];
int pree[MAXM];
int top;
int n, m, s, t, siz, k;
edge()
{
memset(st, -1, sizeof(st));
memset(depth, -1, sizeof(depth));
memset(dis, -1, sizeof(dis));
top = 0;
}
void init()
{
memset(st, -1, sizeof(st));
memset(depth, -1, sizeof(depth));
memset(dis, -1, sizeof(dis));
top = 0;
}
void build(int ts, int tt)
{
s = ts, t = tt, siz = tt;
n = tn; m = tm;
for (int i = 1; i <= n; i++)
add(s, i, p[i], 0);
for (int i = 1; i <= n; i++)
{
for (int j1 = 1; j1 <= m; j1++)
for (int k = 1; k <= sum; k++)
{
add(i, n + (j1 - 1) * sum + k, 1, a[i][j1] * k);
}
}
for (int i = n + 1; i < t; i++)
add(i, t, 1, 0);
}
bool dep()
{
queue<int> q;
q.push(s);
memset(depth, -1, sizeof(depth));
depth[s] = 0;
while (!q.empty())
{
int v = q.front(); q.pop();
for (int i = st[v]; i != -1; i = arr[i].nt)
{
int to = arr[i].to;
if (!arr[i].cal)
continue;
if (depth[to] != -1)
continue;
depth[to] = depth[v] + 1;
q.push(to);
}
}
return (depth[t] != -1);
}
void add(int x, int y, ll z, ll c)
{
top++; arr[top] = { y,st[x],top + 1,c,z }; st[x] = top;
top++; arr[top] = { x,st[y],top - 1,-c,0 }; st[y] = top;
}
ll dfs(int now, ll val)
{
if (now == t || !val)
return val;
ll flow = 0;
for (int& i = cur[now]; i != -1; i = arr[i].nt)
{
int to = arr[i].to;
if (depth[to] != depth[now] + 1)
continue;
ll f = dfs(to, min(arr[i].cal, val));
if (!f || !arr[i].cal)
continue;
flow += f;
arr[i].cal -= f;
arr[arr[i].rev].cal += f;
val -= f;
if (!val)
return flow;
}
return flow;
}
ll dinic()
{
ll flow = 0;
ll f;
while (dep())
{
for (int i = 0; i <= siz; i++)
//这里
cur[i] = st[i];
while (f = dfs(s, INF))
flow += f;
}
return flow;
}
ll min_cost(ll f)
{
ll res = 0;
while (f > 0)
{
fill(dis, dis + 1 + siz, INF);
priority_queue<P, vector<P>, greater<P>> q;
dis[s] = 0;
q.push(P(0, s));
while (!q.empty())
{
P p = q.top(); q.pop();
int v = p.second;
if (dis[v] < p.first) continue;
for (int i = st[v]; i != -1; i = arr[i].nt)
{
note& e = arr[i];
if (e.cal > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to])
{
dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];
pre[e.to] = v;
pree[e.to] = i;
q.push(P(dis[e.to], e.to));
}
}
}
if (dis[t] == INF) return -1;
for (int i = 0; i <= siz; i++)
//这里改了
h[i] += dis[i];
ll d = f;
//printf("2\n");
for (int i = t; i != s; i = pre[i])
{
d = min(d, arr[pree[i]].cal);
}
f -= d;
res += d * h[t];
for (int i = t; i != s; i = pre[i])
{
arr[pree[i]].cal -= d;;
int rev = arr[pree[i]].rev;
arr[rev].cal += d;
}
//printf("3\n");
}
return res;
}
};
edge road, new_road;
edge tmp;
int main()
{
read();
road.init();
road.build(0, tn + sum * tm + 1);
printf("%d", road.min_cost(sum));
return 0;
}【动态加边】
先思考每次遍历的时候会用到哪些边
找一次增广路最多只能找到一条流量为1的增广路
（因为最后连接终点的边流量均为1）
所以说最开始直接把所有的点和边都建好会消耗大量时间
所以我们最初只建完成倒数第一件食物的那一层边
将寻找增广路的过程拆开来看，观察每次查找到增广路之后
被更新的那条边就已经被使用过了，然后将相对应的边加进去
也不是那条边，而是那条边对应的那名厨师，在他的后面加边
说明这名厨师已经在做一道菜了，如果想让他做另外一道菜
就需要排在下一位了
// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cstring>
#define ll int
using namespace std;
const int MAXM = 113234;
const int INF = 2e9 + 7;
typedef pair<int, int> P;
struct note
{
int to;
int nt;
int rev;
ll cost;
ll cal;
};
int tn, tm, tk, sum;
int p[MAXM], a[200][200];
int ha[MAXM], co[MAXM];
void read()
{
scanf("%d%d", &tn, &tm);
for (int i = 1; i <= tn; i++)
scanf("%d", &p[i]), sum += p[i];
for (int i = 1; i <= tn; i++)
for (int j = 1; j <= tm; j++)
scanf("%d", &a[i][j]);
}
struct edge
{
note arr[7000000];
int st[MAXM];
ll dis[MAXM];
ll h[MAXM];
int cur[MAXM];
int depth[MAXM];
int pre[MAXM];
int pree[MAXM];
int top;
int n, m, s, t, siz, k;
edge()
{
memset(st, -1, sizeof(st));
memset(depth, -1, sizeof(depth));
memset(dis, -1, sizeof(dis));
top = 0;
}
void init()
{
memset(st, -1, sizeof(st));
memset(depth, -1, sizeof(depth));
memset(dis, -1, sizeof(dis));
top = 0;
}
void build(int ts, int tt)
{
n = tn; m = tm; s = ts, t = sum*m+n+1, siz = sum*m+n+1;
for (int i = 1; i <= n; i++)
add(s, i+sum*m, p[i], 0);
s = 0, t = sum * m + n + 1;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
add(i + sum * m, (j - 1) * sum + 1, 1, a[i][j]);
}
}
for (int i = 1; i <= m; i++)
add((i - 1) * sum + 1, t, 1, 0);
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= sum; j++)
{
int tmp = (i - 1) * sum + j;
ha[tmp] = j; co[tmp] = i;
}
}
}
bool dep()
{
queue<int> q;
q.push(s);
memset(depth, -1, sizeof(depth));
depth[s] = 0;
while (!q.empty())
{
int v = q.front(); q.pop();
for (int i = st[v]; i != -1; i = arr[i].nt)
{
int to = arr[i].to;
if (!arr[i].cal)
continue;
if (depth[to] != -1)
continue;
depth[to] = depth[v] + 1;
q.push(to);
}
}
return (depth[t] != -1);
}
void add(int x, int y, ll z, ll c)
{
top++; arr[top] = { y,st[x],top + 1,c,z }; st[x] = top;
top++; arr[top] = { x,st[y],top - 1,-c,0 }; st[y] = top;
}
ll dfs(int now, ll val)
{
if (now == t || !val)
return val;
ll flow = 0;
for (int& i = cur[now]; i != -1; i = arr[i].nt)
{
int to = arr[i].to;
if (depth[to] != depth[now] + 1)
continue;
ll f = dfs(to, min(arr[i].cal, val));
if (!f || !arr[i].cal)
continue;
flow += f;
arr[i].cal -= f;
arr[arr[i].rev].cal += f;
val -= f;
if (!val)
return flow;
}
return flow;
}
ll dinic()
{
ll flow = 0;
ll f;
while (dep())
{
for (int i = 0; i <= siz; i++)
//这里
cur[i] = st[i];
while (f = dfs(s, INF))
flow += f;
}
return flow;
}
ll min_cost(ll f)
{
ll res = 0;
fill(dis, dis + 1 + siz, INF);
fill(h, h+1+siz, 0);
priority_queue<P, vector<P>, greater<P>> q;
dis[s] = 0;
q.push(P(0, s));
while (!q.empty())
{
P p = q.top(); q.pop();
int v = p.second;
if (dis[v] < p.first) continue;
for (int i = st[v]; i != -1; i = arr[i].nt)
{
note& e = arr[i];
if (e.cal > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to])
{
dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];
pre[e.to] = v;
pree[e.to] = i;
q.push(P(dis[e.to], e.to));
}
}
}
if (dis[t] == INF) return -1;
for (int i = 0; i <= siz; i++)
//这里改了
h[i] += dis[i];
ll d = f;
//printf("2\n");
for (int i = t; i != s; i = pre[i])
{
d = min(d, arr[pree[i]].cal);
arr[pree[i]].cal -= d;;
int rev = arr[pree[i]].rev;
arr[rev].cal += d;
}
res += d * h[t];
//printf("3\n");
return res;
}
};
edge road, new_road;
edge tmp;
int main()
{
read();
road.init();
road.build(0, tn + sum * tm + 1);
int ret = 0,ans=0;
int n = tn, m = tm;
while(1)
{
ret = road.min_cost(1);
if (ret == -1)
break;
ans += ret;
int pos = road.arr[road.pree[road.t]].rev;
int tmp = road.arr[pos].to;
road.add(tmp + 1, road.t, 1, 0);
for (int i = 1; i <= n; i++)
{
road.add(i + m * sum, tmp + 1, 1, a[i][co[tmp]] * (ha[tmp] + 1));
}
}
printf("%d", ans);
return 0;
}标签：arr,int,luogu,ll,depth,NOI2012,top,美食节,dis 来源： https://www.cnblogs.com/rentu/p/12498707.html