UVA 1161 - Objective: Berlin（最大流）

题目链接：点击打开链接

思路：

一看这些约束条件和数据量， 我们就想到了网络流。

由于有时间， 我们可以构造二元组(u, t)表示在城市u，时间t这个状态，  以这样的二元组作为结点跑最大流， 可惜这样结点高达150*24*60， 会TLE。

我们可以考虑枚举任意两个航线， 如果满足关系， 就建边， 跑最大流。

考虑到每个航线只能经过一次， 我们把航线当作结点， 拆点跑最大流即可。

细节参见代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const double PI = acos(-1);const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18);const int maxn = 10000 + 10;int T,n,m;struct Edge {  int from, to, cap, flow;};bool operator < (const Edge& a, const Edge& b) {  return a.from < b.from || (a.from == b.from && a.to < b.to);}struct Dinic {  int n, m, s, t;        // 结点数， 边数（包括反向弧）， 源点编号， 汇点编号  vector<Edge> edges;    // 边表， edges[e]和edges[e^1]互为反向弧  vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号  bool vis[maxn];        // BFS使用  int d[maxn];           // 从起点到i的距离  int cur[maxn];         // 当前弧指针void init(int n) {    for(int i = 0; i < n; i++) G[i].clear();    edges.clear();}void AddEdge(int from, int to, int cap) {    edges.push_back((Edge){from, to, cap, 0});    edges.push_back((Edge){to, from, 0, 0});    m = edges.size();    G[from].push_back(m-2);    G[to].push_back(m-1);}bool BFS() {    memset(vis, 0, sizeof(vis));    queue<int> Q;    Q.push(s);    vis[s] = 1;    d[s] = 0;    while(!Q.empty()) {      int x = Q.front(); Q.pop();      for(int i = 0; i < G[x].size(); i++) {        Edge& e = edges[G[x][i]];        if(!vis[e.to] && e.cap > e.flow) {  //只考虑残量网络中的弧          vis[e.to] = 1;          d[e.to] = d[x] + 1;          Q.push(e.to);        }      }    }    return vis[t];}int DFS(int x, int a) {    if(x == t || a == 0) return a;    int flow = 0, f;    for(int& i = cur[x]; i < G[x].size(); i++) {  //上次考虑的弧      Edge& e = edges[G[x][i]];      if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {        e.flow += f;        edges[G[x][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()) {      memset(cur, 0, sizeof(cur));      flow += DFS(s, INF);    }    return flow;  }}g;int u, v, c, kase = 0, t, num[maxn], t1[maxn], t2[maxn], a[maxn], b[maxn];char s1[11], s2[11], s3[maxn][10], s4[maxn][10];map<string, int> p;int main() {    while(~scanf("%d", &n)) {        scanf("%s", s1); p.clear();        scanf("%s", s2); kase = 0;        p[s1] = ++kase;  if(!p.count(s2)) p[s2] = ++kase;        int start = p[s1], endd = p[s2];        scanf("%s%d", s1, &m);        t = ((s1[0]-'0')*10+s1[1]-'0')*60+(s1[2]-'0')*10+s1[3]-'0';        for(int i = 1; i <= m; i++) {            scanf("%s%s%d%s%s", s3[i], s4[i], &num[i], s1, s2);            t1[i] = ((s1[0]-'0')*10+s1[1]-'0')*60+(s1[2]-'0')*10+s1[3]-'0';            t2[i] = ((s2[0]-'0')*10+s2[1]-'0')*60+(s2[2]-'0')*10+s2[3]-'0';            if(!p.count(s3[i])) p[s3[i]] = ++kase;            if(!p.count(s4[i])) p[s4[i]] = ++kase;            a[i] = p[s3[i]]; b[i] = p[s4[i]];        }        g.init(2*m + 5);        int res = m;        int src = 2*m + 1, stc = 2*m + 2;        for(int i = 1; i <= m; i++) {            int id1 = b[i];            int cur = a[i];            g.AddEdge(i, i+m, num[i]);            if(start == cur) g.AddEdge(src, i, INF);            if(id1 == endd && t2[i] <= t) g.AddEdge(m+i, stc, INF);            for(int j = 1; j <= m; j++) {                if(i == j) continue;                int id2 = a[j];                if(id1 != id2) continue;                if(t2[i] + 30 <= t1[j]) g.AddEdge(i+m, j, INF);            }        }        printf("%d\n", g.Maxflow(src, stc));    }    return 0;}