标签:int class +++ down res using bitset gcd empty
题意:输入 n 个城市 m 条边,但是边有三种有向边 a b c d,第一种是 d 是 0,那么就是一条普通的路,可以通过无穷多人,如果 d < 0,那么就是隧道,这个隧道是可以藏 c 个人,当然也是通过无穷多人的,如果 d > 0,那么这是一座桥,第一次可以通过一个人,如果修复的话,就可以通过无穷多人,问你最多藏的人数,还有最少花费。
析:只是这样是不能做的,但是题目说了桥不超过 12 个,说实话这个条件太隐蔽了,就是不想让人发现,可惜的是队友没读出来,我也实在是没想出来怎么做,后来一查题解,知道有这个条件,那么很简单了,枚举桥的每一个状态,是修还是不修,每次跑一次最大流,进去判断,下面说一下怎么建图。
建立一个超级源点 s 和超级汇点 t,对于每个城市,从 s 向每个城市连一条边,容量就是城市人数,然后对于普通的路,那么就直接连接容量无穷大,对于隧道也是直接连接容量无穷大,然后把隧道向汇点 t 连接,容量是可以藏人的数,注意连的隧道的左端点,最后是桥,每次枚举桥的状态,如果是修复,那么就连一条容量无穷大的,如果不修复,那么就连一条容量为 1 的边,然后就OK了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define be begin() #define ed end() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x //#define all 1,n,1 #define FOR(i,n,x) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.in", "r", stdin) #define freopenw freopen("out.out", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 100 + 20; const int maxm = 1e6 + 10; const LL mod = 1000000000000000LL; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } inline int readInt(){ int x; scanf("%d", &x); return x; } struct Edge{ int from, to, cap, flow; }; struct Dinic{ int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; void init(int n){ FOR(i, n, 0) G[i].cl; edges.cl; } void addEdge(int from, int to, int cap){ edges.pb((Edge){from, to, cap,0}); edges.pb((Edge){to, from, 0, 0}); m = edges.sz; G[from].pb(m - 2); G[to].pb(m - 1); } bool bfs(){ ms(vis, 0); vis[s] = 1; d[s] = 0; queue<int> q; q.push(s); while(!q.empty()){ int u = q.front(); q.pop(); for(int i = 0; i < G[u].sz; ++i){ Edge &e = edges[G[u][i]]; if(!vis[e.to] && e.cap > e.flow){ d[e.to] = d[u] + 1; vis[e.to] = 1; q.push(e.to); } } } return vis[t]; } int dfs(int u, int a){ if(u == t || a == 0) return a; int flow = 0, f; for(int &i = cur[u]; i < G[u].sz; ++i){ Edge &e = edges[G[u][i]]; if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){ e.flow += f; edges[G[u][i]^1].flow -= f; flow += f; a -= f; if(a == 0) break; } } return flow; } int maxFlow(int s, int t){ this->s = s; this->t = t; int flow = 0; while(bfs()){ ms(cur, 0); flow += dfs(s, INF); } return flow; } }; Dinic dinic; struct Node{ int u, v, c; }; vector<Node> bridge; int main(){ while(scanf("%d %d", &n, &m) == 2){ int s = 0, t = n + 1; dinic.init(t + 5); bridge.cl; for(int i = 1; i <= n; ++i) dinic.addEdge(s, i, readInt()); bool ok = false; for(int i = 1; i <= m; ++i){ int a, b, c, d; scanf("%d %d %d %d", &a, &b, &c, &d); if(d == 0) dinic.addEdge(a, b, INF); else if(d < 0){ dinic.addEdge(a, b, INF); dinic.addEdge(a, t, c); ok = true; } else bridge.pb((Node){a, b, c}); } if(!ok){ puts("Poor Heaven Empire"); continue; } int ans1 = 0, ans2 = 0; int all = 1<<bridge.sz; for(int i = 0; i < all; ++i){ int tmp = 0; for(int j = 0; j < bridge.sz; ++j) if(i&1<<j){ tmp += bridge[j].c; dinic.addEdge(bridge[j].u, bridge[j].v, INF); } else dinic.addEdge(bridge[j].u, bridge[j].v, 1); int res = dinic.maxFlow(s, t); if(res > ans1){ ans1 = res; ans2 = tmp; } else if(res == ans1 && ans2 > tmp) ans2 = tmp; for(int j = 0; j < bridge.sz; ++j){ dinic.edges.pop_back(); dinic.G[bridge[j].u].pop_back(); dinic.G[bridge[j].v].pop_back(); } for(int i = 0; i < dinic.edges.sz; ++i) dinic.edges[i].flow = 0; } if(ans1 == 0) puts("Poor Heaven Empire"); else printf("%d %d\n", ans1, ans2); } return 0; }
HDU 4309 Seikimatsu Occult Tonneru (状压 + 网络流)
标签:int class +++ down res using bitset gcd empty
原文地址:https://www.cnblogs.com/dwtfukgv/p/8972229.html