酒店之王 最大流
题目描述
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?
输入输出格式
输入格式: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。
输入输出样例
输入样例#1:
复制
2 2 2 1 0 1 0 1 1 1 1
输出样例#1: 复制
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; }
EPFL - Fighting