In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships.
You have a transportation company there. Some routes are opened for passengers. Each route is a straight line connecting two different islands, and it is bidirectional. Within an hour, a route can transport a certain number of passengers in one direction. For safety, no two routes are cross or overlap and no routes will pass an island except the departing island and the arriving island. Each island can be treated as a point on the XY plane coordinate system. X coordinate increase from west to east, and Y coordinate increase from south to north.
The transport capacity is important to you. Suppose many passengers depart from the westernmost island and would like to arrive at the easternmost island, the maximum number of passengers arrive at the latter within every hour is the transport capacity. Please calculate it.
题意：给出若干个点和点之间边的流量，问最西边的点到最东边的点的运量是多少
题意描述就是一个裸的网络流最大流问题，直接跑dinic就可以了
1 #pragma comment(linker,"/STACK:16777216")
2 #include<stdio.h>
3 #include<string.h>
4 const int maxm=100100;
5 const int maxv=200100;
6 const int INF=0x3f3f3f3f;
7
8 int s,t;
9 int n,m;
10 int d[maxm],cur[maxm];
11 bool vis[maxm];
12 int head[maxm],point[maxv],flow[maxv],nxt[maxv],size;
13
14 void init(){
15 size=0;
16 memset(head,-1,sizeof(head));
17 }
18
19 void add(int a,int b,int c){
20 point[size]=b;
21 flow[size]=c;
22 nxt[size]=head[a];
23 head[a]=size++;
24 point[size]=a;
25 flow[size]=c;
26 nxt[size]=head[b];
27 head[b]=size++;
28 }
29
30 bool bfs(){
31 memset(vis,0,sizeof(vis));
32 int q[maxm],cnt=0;
33 q[++cnt]=s;
34 vis[s]=1;
35 d[s]=0;
36 for(int i=1;i<=cnt;i++){
37 int u=q[i];
38 for(int j=head[u];~j;j=nxt[j]){
39 if(!vis[point[j]]&&flow[j]>0){
40 d[point[j]]=d[u]+1;
41 q[++cnt]=point[j];
42 vis[point[j]]=1;
43 }
44 }
45 }
46 return vis[t];
47 }
48
49 int dfs(int x,int a){
50 if(x==t||a==0)return a;
51 int ans=0,f;
52 for(int i=head[x];~i;i=nxt[i]){
53 if(d[point[i]]==d[x]+1&&flow[i]>0){
54 f=dfs(point[i],a<flow[i]?a:flow[i]);
55 flow[i]-=f;
56 flow[i^1]+=f;
57 ans+=f;
58 a-=f;
59 if(a==0)break;
60 }
61 }
62 if(ans==0)d[x]=-1;
63 return ans;
64 }
65
66 int mf(){
67 int ans=0;
68 while(bfs()){
69 ans+=dfs(s,INF);
70 }
71 return ans;
72 }
73
74 int main(){
75 int T;
76 scanf("%d",&T);
77 for(int q=1;q<=T;q++){
78
79 init();
80 scanf("%d%d",&n,&m);
81 int i,j,minx,maxx;
82 for(i=1;i<=n;i++){
83 int x,y;
84 scanf("%d%d",&x,&y);
85 if(i==1){
86 minx=x;
87 maxx=x;
88 s=t=i;
89 }
90 else{
91 if(x>maxx){
92 maxx=x;
93 s=i;
94 }
95 else if(x<minx){
96 minx=x;
97 t=i;
98 }
99 }
100 }
101 for(i=1;i<=m;i++){
102 int a,b,v;
103 scanf("%d%d%d",&a,&b,&v);
104 add(a,b,v);
105 }
106 printf("%d\n",mf());
107 }
108 return 0;
109 }
