SGU 438 The Glorious Karlutka River =) 动态网络流

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题意：给你一条宽为w的河河中央有一些石头有一些人在河的一面它们想到河的另一面每个人的弹跳距离为d 每个石头每一时刻有一个最大载人数量问你所有人是否能跳到河对面如果能输出最小时间

思路：这道题并不是我们常做的静态网络流因为每一秒中网络中的流量可能不相同所以我们要用动态的思想

常见的动态网络流的题型为固定时间求最大通过流量和固定流量求最小通过时间显然这道题属于第二种

建图方面我们按时间拆点因为如果可以到达河对岸那么最长时间一定小于等于n+m 所以我们把原图拆成n+m个点对于第i秒的石头我们把第i-1秒的该石头连一条边指向它代表在第i-1秒的人在这个石头上待了一秒没动然后上一秒的石头a与它相连的下一秒的石头b相连代表上一秒在a石头上的人花了一秒钟跳到了b石头上然后源点与这一秒的石头相连代表在岸边的人花了一秒钟的时间跳到第一个石头上上一秒的石头与对岸相连为什么是上一秒的石头呢？因为是那个人在第i-1秒花了一秒在第i秒才跳到了对岸然后每一秒中跑最大流记录最大流流量即可

这道题我与到一个问题就是在windows上运行这个程序输入 0 5 2 1 这组数据它会蹦掉但是我如果加上中间输出它就莫名的不蹦了... 哪位大神知道其中的原因一定要帮我解答感激不尽 T T

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <vector>using namespace std;#define REP( i, a, b ) for( int i = a; i < b; i++ )#define FOR( i, a, b ) for( int i = a; i <= b; i++ )#define CLR( a, x ) memset( a, x, sizeof a )#define CPY( a, x ) memcpy( a, x, sizeof a )const int maxn = 10000 + 1000;const int maxe = 600000 + 10;const int INF = 1e9;struct Edge{          int v, c, f;          int next;          Edge() {}          Edge(int v, int c, int f, int next) : v(v), c(c), f(f), next(next) {}};struct ISAP{          int n, s, t;          int num[maxn], cur[maxn], d[maxn], p[maxn];          int Head[maxn], cntE;          int Q[maxn], head, tail;          Edge edge[maxe];          void Init(int n){                    this -> n = n;                    cntE = 0;                    CLR(Head, -1);          }          void Add(int u, int v, int c){                    edge[cntE] = Edge(v, c, 0, Head[u]);                    Head[u] = cntE++;                    edge[cntE] = Edge(u, 0, 0, Head[v]);                    Head[v] = cntE++;          }          void Bfs(){                    CLR(d, -1);                    CLR(num, 0);                    d[t] = 0;                    head = tail = 0;                    Q[tail++] = t;                    num[0] = 1;                    while(head != tail){                              int u = Q[head++];                              for(int i = Head[u]; ~i; i = edge[i].next){                                        Edge &e = edge[i];                                        if(~d[e.v]) continue;                                        d[e.v] = d[u] + 1;                                        Q[tail++] = e.v;                                        num[d[e.v]] ++;                              }                    }          }          int Maxflow(int s, int t){                    this -> s = s;                    this -> t = t;                    CPY(cur, Head);                    Bfs();                    int flow = 0, u = p[s] = s;                    while(d[s] < n){                              if(u == t){                                        int f = INF, neck;                                        for(int i = s; i != t; i = edge[cur[i]].v){                                                  if(f > edge[cur[i]].c - edge[cur[i]].f){                                                            f = edge[cur[i]].c - edge[cur[i]].f;                                                            neck = i;                                                  }                                        }                                        for(int i = s; i != t; i = edge[cur[i]].v){                                                  edge[cur[i]].f += f;                                                  edge[cur[i]^1].f -= f;                                        }                                        flow += f;                                        u = neck;                              }                              int ok = 0;                              for(int i = cur[u]; ~i; i = edge[i].next){                                        Edge &e = edge[i];                                        if(e.c > e.f && d[e.v] + 1 == d[u]){                                                  ok = 1;                                                  cur[u] = i;                                                  p[e.v] = u;                                                  u = e.v;                                                  break;                                        }                              }                              if(!ok){                                        int m = n - 1;                                        if(--num[d[u]] == 0) break;                                        for(int i = Head[u]; ~i; i = edge[i].next){                                                  Edge &e = edge[i];                                                  if(e.c - e.f > 0 && m > d[e.v]){                                                            cur[u] = i;                                                            m = d[e.v];                                                  }                                        }                                        ++num[d[u] = m + 1];                                        u = p[u];                              }                    }                    return flow;          }}solver;const double eps = 1e-8;int dcmp(double x){    if(fabs(x) < eps) return 0;    return x < 0 ? -1 : 1;}struct Point{    double x, y;    int c;    void read(){        scanf("%lf%lf%d", &x, &y, &c);    }    bool can(Point a, double d){        return dcmp((x - a.x) * (x - a.x) + (y - a.y) * (y - a.y) - d * d) <= 0;    }}p[60];int n, m;int st, ed;double d, w;vector<int> G[60];int in(int T, int u){ return 2 * T * n + u; }int out(int T, int u){ return (2 * T + 1) * n + u; }void build_relationship(){    FOR(i, 1, n) FOR(j, i+1, n) if(p[i].can(p[j], d)){        G[i].push_back(j);        G[j].push_back(i);    }    FOR(i, 1, n) if(p[i].y <= d) G[st].push_back(i);}void build_netflow(int t){    FOR(i, 1, n) solver.Add(in(t, i), out(t, i), p[i].c);    REP(i, 0, (int)G[st].size()){        int tar = G[st][i];        solver.Add(st, in(t, tar), INF);    }    if(w <= d) solver.Add(st, ed, INF);    FOR(i, 1, n) REP(j, 0, (int)G[i].size()){        int tar = G[i][j];        solver.Add(out(t-1, i), in(t, tar), INF);    }    FOR(i, 1, n) if(p[i].y + d >= w) solver.Add(out(t-1, i), ed, INF);}void solve(){    scanf("%d%d%lf%lf", &n, &m, &d, &w);    st = 0, ed = (n + m + 1) * n * 2 + 1;    solver.Init(ed + 1);    //printf("....\n");    FOR(i, 1, n) p[i].read();    build_relationship();    if(m == 0){          printf("0\n");          return;    }    int sum = 0, ans = 0;    FOR(i, 1, m + n){        build_netflow(i);        sum += solver.Maxflow(st, ed);        if(sum >= m){ ans = i; break; }    }    if(sum >= m) printf("%d\n", ans);    else printf("IMPOSSIBLE\n");}int main(){    solve();    return 0;}/*0 5 2 1*/

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