hdu 1532

心情不爽找到水题刷刷，很裸的网络流！直接模版过！！！！

#include <stdio.h>
#include <string.h>
#include<iostream>
using namespace std;
const int VM = 220, EM = 20500, INF = 0x3f3f3f3f;
struct E
{
    int to, frm, nxt, cap;
}edge[EM];
int head[VM], e, n, src, des;
int dep[VM], gap[VM]; //gap[x]=y:说明残留网络中dep[i]=x的个数为y
//dep记录到汇点的距离，gap[i]记录到汇点的距离为i的个数
void 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++;
}

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 Q[VM], front = 0, rear = 0;
    dep[des] = 0;
    Q[rear++] = des;
    int u, v;
    while (front != rear)
    {
        u = Q[front++];
        front = front%VM;
        for (int i=head[u]; i!=-1; i=edge[i].nxt)
        {
            v = edge[i].to;
            if (edge[i].cap != 0 || dep[v] != -1)
                continue;
            Q[rear++] = v;
            rear = rear % VM;
            ++gap[dep[v] = dep[u] + 1];  //求出各层次的数量
        }
    }
}

int SAP(int src, int des)
{
    int res = 0;
    BFS(src, des);
    int cur[VM];
    int S[VM], top = 0;
    memcpy(cur, head, sizeof(head));
    int u = src, i;
    while (dep[src] < n)   //n为结点的个数
    {
        if (u == des)
        {
            int temp = INF, inser = n;
            for (i=0; i!=top; ++i)
                if (temp > 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 m,num;
    while (scanf("%d%d", &m,&n) != EOF)
    {
        e = 0;
        memset(head, -1, sizeof(head));
        int u, v, c, src =1, des =n;
        int temp=n;
        for(int i=0;i<m;i++)
        {
            cin>>u>>v>>c;
            addedge(u,v,c);
        }
        printf("%d\n", SAP(src, des));
    }
    return 0;
}