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;}