【47.63%】【hdu 1532】Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15777 Accepted Submission(s): 7514

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
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output
50

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1532

【题解】

让你找从点1到点m的最大流;
用那个E-K算法搞增广路做就好;
用广搜来搞;
路上的边权是剩余网络的边权;
每次修改a后;
正向边+a;
反向边-a;
一开始输入的z是这个网络上边的容量限制(一开始没有任何流);
所以每次修改a后就直接增加答案a;
中途跳出队列可能dl不为空，所以要清空队列；
如果还能找到增广路就继续找;继续增加最大流;

【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 300;
const int INF = 2100000000;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

int n,m;
int pre[MAXN];
int flow[MAXN][MAXN];
bool mark[MAXN];
vector <int> a[MAXN];
queue <int>dl;

void read2(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void read1(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}


int main()
{
    //freopen("F:\rush.txt","r",stdin);
    while (~scanf("%d%d",&n,&m))
    {
        memset(flow,0,sizeof(flow));
        for (int i = 1;i <= n;i++)
        {
            int x,y,z;
            read1(x);read1(y);read1(z);
            flow[x][y]+=z;
        }
        int f = 0;
        while (true)
        {
            memset(mark,false,sizeof(mark));
            memset(pre,0,sizeof(pre));
            while (!dl.empty())
                dl.pop();
            dl.push(1);
            mark[1] = true;
            while (!dl.empty())
            {
                int x = dl.front();
                dl.pop();
                if (x==m)
                    break;
                for (int i = 1;i <= m;i++)
                    if (flow[x][i]>0 && !mark[i])
                    {
                        mark[i] = true;
                        pre[i] = x;
                        dl.push(i);
                    }
            }
            if (!mark[m])
                break;
            int mi = INF;
            int i = m;
            while (i!=1)
            {
                mi = min(mi,flow[pre[i]][i]);
                i = pre[i];
            }
            if (mi==INF)
                break;
            i = m;
            while (i!=1)
            {
                flow[pre[i]][i]-=mi;
                flow[i][pre[i]]+=mi;
                i = pre[i];
            }
            f+=mi;
        }
        cout << f<<endl;
    }
    return 0;
}