ZOJ 2760 How Many Shortest Path 最大流+Floyed

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题意：给你一个带权有向图问你起点到终点距离最短且边不重复的路径有多少条
思路：思路很巧秒先Floyed处理出任意两点的最短路径然后如果floyed[S][i] + dist[i][j] + floyed[j][T] == floyed[S][T] && dist[i][j] != INF 则保存i j 这条边容量为1 代表只能用一次
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define REP( 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 = 100 + 10;const int maxe = 20000 + 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;int n, S, T;int dist[maxn][maxn], floyed[maxn][maxn];void input(){          REP(i, 0, n) REP(j, 0, n){                    scanf("%d", &dist[i][j]);                    if(dist[i][j] == -1) dist[i][j] = INF;                    if(i == j) dist[i][j] = 0;//太坑了 不加这句就WA到死... 有意思么...                    floyed[i][j] = dist[i][j];          }          scanf("%d%d", &S, &T);}void solve(){          if(S == T){                    printf("inf\n");                    return;          }          REP(k, 0, n) REP(i, 0, n) REP(j, 0, n)                    if(floyed[i][k] != INF && floyed[k][j] != INF)                              floyed[i][j] = min(floyed[i][j], floyed[i][k] + floyed[k][j]);          if(floyed[S][T] == INF){                    printf("0\n");                    return;          }          solver.Init(n);          REP(i, 0, n) REP(j, 0, n)                    if(floyed[S][i] + dist[i][j] + floyed[j][T] == floyed[S][T] && dist[i][j] != INF)                              solver.Add(i, j, 1);          printf("%d\n", solver.Maxflow(S, T));}int main(){          //freopen("in.txt", "r", stdin);          while(~scanf("%d", &n)){                    input();                    solve();          }          return 0;}
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