POJ1273&&Hdu1532 Drainage Ditches（最大流dinic）

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本文探讨了如何解决农场中雨水通过排水沟渠系统流向溪流的问题，利用最大流算法来确定从池塘到溪流的最大排水率。通过具体实例展示了如何使用Dinic算法进行求解。

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16314 Accepted Submission(s): 7748

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

Sample Output

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题目求从1到n的最大水流量，就是求最大流很显然的板子题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500

struct node
{
    int u, v, next, cap;
} edge[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
int cnt;

void init()
{
    cnt = 0;
    memset(s, -1, sizeof(s));
}

void add(int u, int v, int c)
{
    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].cap = c;
    edge[cnt].next = s[u];
    s[u] = cnt++;
    edge[cnt].u = v;
    edge[cnt].v = u;
    edge[cnt].cap = 0;
    edge[cnt].next = s[v];
    s[v] = cnt++;
}

bool BFS(int ss, int ee)
{
    memset(d, 0, sizeof d);
    d[ss] = 1;
    queue<int>q;
    q.push(ss);
    while (!q.empty())
    {
        int pre = q.front();
        q.pop();
        for (int i = s[pre]; ~i; i = edge[i].next)
        {
            int v = edge[i].v;
            if (edge[i].cap > 0 && !d[v])
            {
                d[v] = d[pre] + 1;
                q.push(v);
            }
        }
    }
    return d[ee];
}

int DFS(int x, int exp, int ee)
{
    if (x == ee||!exp) return exp;
    int temp,flow=0;
    for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
    {
        int v = edge[i].v;
        if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
        {
            edge[i].cap -= temp;
            edge[i ^ 1].cap += temp;
            flow += temp;
            exp -= temp;
            if (!exp) break;
        }
    }
    if (!flow) d[x] = 0;
    return flow;
}

int Dinic_flow(int ss, int ee)
{
    int ans = 0;
    while (BFS(ss, ee))
    {
        for (int i = 0; i <= ee; i++) nt[i] = s[i];
        ans+= DFS(ss, INF, ee);
    }
    return ans;
}

int main()
{
    int n,m,u,v,c;
    while (~scanf("%d %d", &m, &n))
    {
        init();
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            add(u,v,c);
        }
        printf("%d\n",Dinic_flow(1,n));

    }
    return 0;
}