【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：

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cmath>
  5 #include<cstdlib>
  6 #include<string>
  7 #include<queue>
  8 #include<algorithm>
  9 using namespace std;
 10  
 11 const int N=100010;
 12 const int M=400010;
 13 const int inf=0xfffffff;
 14  
 15 int n,m,cnt;
 16  
 17 struct Edge{
 18     int v , cap , next;
 19 } edge[M];
 20  
 21 int head[N],pre[N],d[N],numd[N];
 22 int cur_edge[N];
 23 
 24 void addedge(int u,int v,int c){
 25     edge[cnt].v = v;
 26     edge[cnt].cap = c;
 27     edge[cnt].next = head[u];
 28     head[u] = cnt++;
 29  
 30     edge[cnt].v = u;
 31     edge[cnt].cap = 0;
 32     edge[cnt].next = head[v];
 33     head[v] = cnt++;
 34 }
 35  
 36 void bfs(int s){
 37     memset(numd,0,sizeof(numd));
 38     for(int i=1; i<=n; i++)    numd[ d[i] = n ]++;
 39     d[s] = 0;
 40     numd[n]--;
 41     numd[0]++;
 42     queue<int> Q;
 43     Q.push(s);
 44  
 45     while(!Q.empty()){
 46         int v=Q.front();
 47         Q.pop();
 48  
 49         int i=head[v];
 50         while(i != -1){
 51             int u=edge[i].v;
 52  
 53             if(d[u]<n){
 54                 i=edge[i].next;
 55                 continue ;
 56             }
 57  
 58             d[u] = d[v]+1;
 59             numd[n]--;
 60             numd[d[u]]++;
 61             Q.push(u);
 62             i=edge[i].next;
 63         }
 64     }
 65 }
 66 
 67 int SAP(int s,int t){
 68     for(int i = 1; i <= n; i++)    cur_edge[i] = head[i];
 69     int max_flow = 0;
 70     bfs(t);
 71     int u = s ;
 72     while(d[s] < n){
 73         if(u == t){
 74             int cur_flow = inf,neck;
 75             for(int from = s; from != t; from = edge[cur_edge[from]].v){
 76                 if(cur_flow > edge[cur_edge[from]].cap){
 77                     neck = from;
 78                     cur_flow = edge[cur_edge[from]].cap;
 79                 }
 80             }
 81  
 82             for(int from = s; from != t; from = edge[cur_edge[from]].v){    //修改增广路上的边的容量
 83                 int tmp = cur_edge[from];
 84                 edge[tmp].cap -= cur_flow;
 85                 edge[tmp^1].cap += cur_flow;
 86             }
 87             max_flow += cur_flow;
 88             u = neck;
 89         }
 90  
 91         int i;
 92         for(i = cur_edge[u]; i != -1; i = edge[i].next)
 93             if(edge[i].cap && d[u] == d[edge[i].v]+1)
 94                 break;
 95  
 96         if(i != -1){
 97             cur_edge[u] = i;
 98             pre[edge[i].v] = u;
 99             u = edge[i].v;
100         }
101         else{
102             numd[d[u]]--;
103             if(!numd[d[u]]) break;  
104             cur_edge[u] = head[u];
105             int tmp = n;
106             for(int j = head[u]; j != -1; j = edge[j].next)
107                 if(edge[j].cap && tmp > d[edge[j].v])
108                     tmp = d[edge[j].v];
109  
110             d[u] = tmp+1;
111             numd[d[u]]++;
112             if(u != s)
113                 u = pre[u];
114         }
115     }
116     return max_flow;
117 }
118 
119 inline void pre_init(){ 
120     cnt = 0;
121     memset(head, -1, sizeof head);  
122 } 
123 
124 void mapping(){
125     int u, v, w;
126     for(int i = 1; i <= m; ++i){  
127         scanf("%d %d %d", &u, &v, &w); 
128         addedge(u, v, w);
129         addedge(v, u, w);
130     }
131 }
132 
133 int main(){
134 
135     int tcase;
136     scanf("%d",&tcase);
137     while(tcase--){
138         scanf("%d%d",&n,&m);
139         int x,y,s,t;
140         int minn = inf, maxx = -inf;
141         for(int i = 1; i <= n; i++){
142             scanf("%d%d",&x,&y);
143             if(x <= minn){
144                 s = i;
145                 minn = x;
146             }
147             if(x >= maxx){
148                 t = i;
149                 maxx = x;
150             }
151         }
152         pre_init();
153         mapping();
154         int ans = SAP(s,t);
155         printf("%d\n",ans);
156     }
157     return 0;
158 }