Island Transport
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 7225 Accepted Submission(s): 2256
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
2
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
9
6
题意：
共有t组数据，对于每组数据，一共有n个小岛，m条线路，且每个小岛都有自己对应的位置。每条线路是从一个小岛直达另一个小岛的线路，每小时能运送c人。
现在要求从最西边（x最小）到最东边（x最大）每小时最多能运送多少人。
题解：
在输入坐标的时候记录一下最西边和最东边的小岛编号作为S和T，然后建网跑最大流即可。
需要注意的是这里的数据比较大，达到了10W，所以要用上优化。（未优化前超时，后3S）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <string>
using namespace std;
#define LL long long int
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{tol = 0;memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{memset(dep,-1,sizeof(dep));memset(gap,0,sizeof(gap));gap[0] = 1;int front = 0, rear = 0;dep[end] = 0;Q[rear++] = end;while(front != rear){int u = Q[front++];for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(dep[v] != -1)continue;Q[rear++] = v;dep[v] = dep[u] + 1;gap[dep[v]]++;}}
}
int S[MAXN];
int sap(int start,int end,int N)
{BFS(start,end);memcpy(cur,head,sizeof(head));int top = 0;int u = start;int ans = 0;while(dep[start] < N){if(u == end){int Min = INF;int inser;for(int i = 0; i < top; i++)if(Min > edge[S[i]].cap - edge[S[i]].flow){Min = edge[S[i]].cap - edge[S[i]].flow;inser = i;}for(int i = 0; i < top; i++){edge[S[i]].flow += Min;edge[S[i]^1].flow -= Min;}ans += Min;top = inser;u = edge[S[top]^1].to;continue;}bool flag = false;int v;for(int i = cur[u]; i != -1; i = edge[i].next){v = edge[i].to;if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){flag = true;cur[u] = i;break;}}if(flag){S[top++] = cur[u];u = v;continue;}int Min = N;for(int i = head[u]; i != -1; i = edge[i].next)if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){Min = dep[edge[i].to];cur[u] = i;}gap[dep[u]]--;if(!gap[dep[u]])return ans;dep[u] = Min + 1;gap[dep[u]]++;if(u != start)u = edge[S[--top]^1].to;}return ans;
}
int t;
LL n,m;
int main()
{LL x,y,S,T,west,east,c;scanf("%d",&t);while(t--){init();west=INF,east=-INF;scanf("%lld%lld",&n,&m);for(LL i=1;i<=n;i++){scanf("%lld%lld",&x,&y);if(x<=west)west=x,S=i;if(x>=east)east=x,T=i;}for(LL i=1;i<=m;i++){scanf("%lld%lld%lld",&x,&y,&c);addedge(x,y,c,c);}cout << sap(S,T,n) << endl;}return 0;
}