hdu4280——Island Transport（最大流SAP算法）-白红宇

hdu4280——Island Transport（最大流SAP算法）

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发布时间：2019-05-10

本文共 4267 字，大约阅读时间需要 14 分钟。

Problem Description

　　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.

Input

　　The first line contains one integer T (1<=T<=20), the number of test cases.

　　Then T test cases follow. The first line of each test case contains two integers N and M (2<=N,M<=100000), the number of islands and the number of routes. Islands are number from 1 to N.

　　Then N lines follow. Each line contain two integers, the X and Y coordinate of an island. The K-th line in the N lines describes the island K. The absolute values of all the coordinates are no more than 100000.

　　Then M lines follow. Each line contains three integers I1, I2 (1<=I1,I2<=N) and C (1<=C<=10000) . It means there is a route connecting island I1 and island I2, and it can transport C passengers in one direction within an hour.

　　It is guaranteed that the routes obey the rules described above. There is only one island is westernmost and only one island is easternmost. No two islands would have the same coordinates. Each island can go to any other island by the routes.

Output

　　For each test case, output an integer in one line, the transport capacity.

Sample Input

5 7

3 3

3 0

3 1

0 0

4 5

1 3 3

2 3 4

2 4 3

1 5 6

4 5 3

1 4 4

3 4 2

6 7

-1 -1

0 1

0 2

1 0

1 1

2 3

1 2 1

2 3 6

4 5 5

5 6 3

1 4 6

2 5 5

3 6 4

Sample Output

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#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #define INF 0x3f3f3f3f#define MAXN 100005#define Mod 10001using namespace std;struct E{ int to, frm, nxt, cap;}edge[MAXN<<2];int head[MAXN], e, n, m, src, des;int dep[MAXN], gap[MAXN]; //gap[x]=y:说明残留网络中dep[i]=x的个数为yvoid addedge(int u, int v, int c){ edge[e].frm = u; edge[e].to = v; edge[e].cap = c; edge[e].nxt = head[u]; head[u] = e++; edge[e].frm = v; edge[e].to = u; edge[e].cap = 0; edge[e].nxt = head[v]; head[v] = e++;}int Q[MAXN];void BFS(int src, int des){ memset(dep, -1, sizeof(dep)); memset(gap, 0, sizeof(gap)); gap[0] = 1; //说明此时有1个dep[i] = 0 int front = 0, rear = 0; dep[des] = 0; Q[rear++] = des; int u, v; while (front != rear) { u = Q[front++]; //cout<
             <
              
                edge[S[i]].cap) { temp = edge[S[i]].cap; inser = i; } for (i=0; i!=top; ++i) { edge[S[i]].cap -= temp; edge[S[i]^1].cap += temp; } res += temp; top = inser; u = edge[S[top]].frm; } if (u != des && gap[dep[u] -1] == 0)//出现断层，无增广路 break; for (i = cur[u]; i != -1; i = edge[i].nxt)//遍历与u相连的未遍历结点 if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1) //层序关系， 找到允许 break; if (i != -1)//找到允许弧 { cur[u] = i; S[top++] = i;//加入路径栈 u = edge[i].to;//查找下一个结点 } else //无允许的路径，修改标号 当前点的标号比与之相连的点中最小的多1 { int min = n; for (i = head[u]; i != -1; i = edge[i].nxt) //找到与u相连的v中dep[v]最小的点 { if (edge[i].cap == 0) continue; if (min > dep[edge[i].to]) { min = dep[edge[i].to]; cur[u] = i; //最小标号就是最新的允许弧 } } --gap[dep[u]]; //dep[u] 的个数变化了 所以修改gap ++gap[dep[u] = min + 1]; //将dep[u]设为min(dep[v]) + 1, 同时修改相应的gap[] if (u != src) //该点非源点&&以u开始的允许弧不存在，退点 u = edge[S[--top]].frm; } } return res;}int main(){ int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); int l=INF,r=-INF; for(int i=1;i<=n;++i) { int x,y; scanf("%d%d",&x,&y); if(x
               
                r) { r=x; des=i; } } e=0; memset(head,-1,sizeof(head)); for(int i=1;i<=m;++i) { int x,y,w; scanf("%d%d%d",&x,&y,&w); addedge(x,y,w); addedge(y,x,w); } int ans=SAP(); printf("%d\n",ans); } return 0;}

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