标签：return set for tin orange const stack scanf val
XX酒店的老板想成为酒店之王，本着这种希望，第一步要将酒店变得人性化。由于很多来住店的旅客有自己喜好的房间色调、阳光等，也有自己所爱的菜，但是该酒店只有p间房间，一天只有固定的q道不同的菜。
有一天来了n个客人，每个客人说出了自己喜欢哪些房间，喜欢哪道菜。但是很不幸，可能做不到让所有顾客满意（满意的条件是住进喜欢的房间，吃到喜欢的菜）。
这里要怎么分配，能使最多顾客满意呢？
第一行给出三个正整数表示n,p,q(<=100)。
之后n行，每行p个数包含0或1，第i个数表示喜不喜欢第i个房间（1表示喜欢，0表示不喜欢）。
之后n行，每行q个数，表示喜不喜欢第i道菜。
输出格式：最大的顾客满意数。
2 2 2 1 0 1 0 1 1 1 1
1
最大流；
建边，将人拆成两个点 p1,p2；
首先由源点向每道菜连容量为1的边，
每个房间向汇点连容量为1的边；
然后每个人的 p1 与他喜欢的菜连容量为1的边，p1与p2连容量为1的边，p2与他喜欢的房间连容量为1的边；
然后 dinic 跑一遍最大流即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int n, p, q;
bool vis[maxn];
int x, y, f, z;
struct node {
int to, nxt, val, u;
}edge[maxn << 2];
int head[maxn], cnt;
void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].to = v; edge[cnt].val = w;
edge[cnt].nxt = head[u]; head[u] = cnt++;
}
int st, ed, rk[maxn];
int bfs() {
queue<int>qq;
ms(rk);
rk[st] = 1;
qq.push(st);
while (!qq.empty()) {
int tmp = qq.front(); qq.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].to;
if (rk[to] || edge[i].val <= 0)continue;
rk[to] = rk[tmp] + 1; qq.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].to;
if (rk[v] != rk[u] + 1 || !edge[i].val)continue;
int tmpadd = dfs(v, min(edge[i].val, flow - add));
if (!tmpadd) {
rk[v] = -1; continue;
}
edge[i].val -= tmpadd; edge[i ^ 1].val += tmpadd;
add += tmpadd;
}
return add;
}
int ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
int main()
{
//ios::sync_with_stdio(0);
memset(head, -1, sizeof(head)); cnt = 0;
rdint(n); rdint(p); rdint(q);
st = p + q + 2 * n + 1; ed = st + 1;
for (int i = 1; i <= p; i++)addedge(st, i, 1), addedge(i, st, 0);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= p; j++) {
int x; rdint(x);
if (x == 1) {
addedge(p + i, j, 0);
addedge(j, p + i, 1);
}
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= q; j++) {
int x; rdint(x);
if (x == 1) {
addedge(p + q + i + n, p + j + n, 1);
addedge(p + j + n, p + q + i + n, 0);
}
}
}
for (int i = p + 1 + n; i <= p + n + q; i++) {
addedge(i, ed, 1); addedge(ed, i, 0);
}
for (int i = 1; i <= n; i++)
addedge(p + i, p + q + i + n, 1), addedge(p + q + i + n, p + i, 0);
dinic();
cout << ans << endl;
return 0;
}
标签：return set for tin orange const stack scanf val
原文地址：https://www.cnblogs.com/zxyqzy/p/9981354.html
