【HDUOJ】4280 Island Transport

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4280

题意：有n个岛屿,m条无向路,每个路给出最大允许的客流量,求从最西的那个岛屿最多能运用多少乘客到最东的那个岛屿

题解：最大流的裸题。输入记得找到最西和最东的岛屿，以及注意是双向边。。这题用的sap.

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<string>
 #include<queue>
 #include<algorithm>
 using namespace std;

 const int N=;
 const int M=;
 const int inf=0xfffffff;

 int n,m,cnt;

 struct Edge{
     int v , cap , next;
 } edge[M];

 int head[N],pre[N],d[N],numd[N];
 int cur_edge[N];

 void addedge(int u,int v,int c){
     edge[cnt].v = v;
     edge[cnt].cap = c;
     edge[cnt].next = head[u];
     head[u] = cnt++;

     edge[cnt].v = u;
     edge[cnt].cap = ;
     edge[cnt].next = head[v];
     head[v] = cnt++;
 }

 void bfs(int s){
     memset(numd,,sizeof(numd));
     for(int i=; i<=n; i++)    numd[ d[i] = n ]++;
     d[s] = ;
     numd[n]--;
     numd[]++;
     queue<int> Q;
     Q.push(s);

     while(!Q.empty()){
         int v=Q.front();
         Q.pop();

         int i=head[v];
         while(i != -){
             int u=edge[i].v;

             if(d[u]<n){
                 i=edge[i].next;
                 continue ;
             }

             d[u] = d[v]+;
             numd[n]--;
             numd[d[u]]++;
             Q.push(u);
             i=edge[i].next;
         }
     }
 }

 int SAP(int s,int t){
     for(int i = ; i <= n; i++)    cur_edge[i] = head[i];
     int max_flow = ;
     bfs(t);
     int u = s ;
     while(d[s] < n){
         if(u == t){
             int cur_flow = inf,neck;
             for(int from = s; from != t; from = edge[cur_edge[from]].v){
                 if(cur_flow > edge[cur_edge[from]].cap){
                     neck = from;
                     cur_flow = edge[cur_edge[from]].cap;
                 }
             }

             for(int from = s; from != t; from = edge[cur_edge[from]].v){    //修改增广路上的边的容量
                 int tmp = cur_edge[from];
                 edge[tmp].cap -= cur_flow;
                 edge[tmp^].cap += cur_flow;
             }
             max_flow += cur_flow;
             u = neck;
         }

         int i;
         for(i = cur_edge[u]; i != -; i = edge[i].next)
             if(edge[i].cap && d[u] == d[edge[i].v]+)
                 break;

         if(i != -){
             cur_edge[u] = i;
             pre[edge[i].v] = u;
             u = edge[i].v;
         }
         else{
             numd[d[u]]--;
             if(!numd[d[u]]) break;
             cur_edge[u] = head[u];
             int tmp = n;
             for(int j = head[u]; j != -; j = edge[j].next)
                 if(edge[j].cap && tmp > d[edge[j].v])
                     tmp = d[edge[j].v];

             d[u] = tmp+;
             numd[d[u]]++;
             if(u != s)
                 u = pre[u];
         }
     }
     return max_flow;
 }

 inline void pre_init(){
     cnt = ;
     memset(head, -, sizeof head);
 } 

 void mapping(){
     int u, v, w;
     for(int i = ; i <= m; ++i){
         scanf("%d %d %d", &u, &v, &w);
         addedge(u, v, w);
         addedge(v, u, w);
     }
 }

 int main(){

     int tcase;
     scanf("%d",&tcase);
     while(tcase--){
         scanf("%d%d",&n,&m);
         int x,y,s,t;
         int minn = inf, maxx = -inf;
         for(int i = ; i <= n; i++){
             scanf("%d%d",&x,&y);
             if(x <= minn){
                 s = i;
                 minn = x;
             }
             if(x >= maxx){
                 t = i;
                 maxx = x;
             }
         }
         pre_init();
         mapping();
         int ans = SAP(s,t);
         printf("%d\n",ans);
     }
     return ;
 }