POJ 2455 Secret Milking Machine 二分+最大流 ISAP + GAP优化

题目要求如下

一个无向图，有一些双向边。起点为1，目标点为n。要求必须有t条不同的路径。所谓的不同的路径，是一条边只能属于一个路径。

现在求所有路径的最大边 最小的长度是多少。

脑海中顿时闪过二分+最大流，找出模板秒之，因为还没到那种闭目写网络流的程度。 这次的模板是ISAP，我将一个大牛写的指针版本的邻接表改成了数组版本的。

然后还有一个技巧是二分的上下界最好用所有边长度的上下界，因为可能有恶心数据卡时限。

我用了输入优化后207ms，没用之前250ms

/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 222
#define MAXM 164444
#define INF 100000000
#define eps 1e-7
using namespace std;
struct node
{
    int ver;    // vertex
    int cap;    // capacity
    int flow;   // current flow in this arc
    int next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e;
int xx[MAXM], yy[MAXM], cc[MAXM];
void add(int x, int y, int c)
{       //e记录边的总数
    edge[e].ver = y;
    edge[e].cap = c;
    edge[e].flow = 0;
    edge[e].rev = e + 1;        //反向边在edge中的下标位置
    edge[e].next = head[x];   //记录以x为起点的上一条边在edge中的下标位置
    head[x] = e++;           //以x为起点的边的位置
    //反向边
    edge[e].ver = x;
    edge[e].cap = 0;  //反向边的初始网络流为0
    edge[e].flow = 0;
    edge[e].rev = e - 1;
    edge[e].next = head[y];
    head[y] = e++;
}
void rev_BFS()
{
    int Q[MAXN], qhead = 0, qtail = 0;
    for(int i = 1; i <= n; ++i)
    {
        dist[i] = MAXN;
        numbs[i] = 0;
    }
    Q[qtail++] = des;
    dist[des] = 0;
    numbs[0] = 1;
    while(qhead != qtail)
    {
        int v = Q[qhead++];
        for(int i = head[v]; i != -1; i = edge[i].next)
        {
            if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;
            dist[edge[i].ver] = dist[v] + 1;
            ++numbs[dist[edge[i].ver]];
            Q[qtail++] = edge[i].ver;
        }
    }
}

int maxflow()
{
    int u, totalflow = 0;
    int Curhead[MAXN], revpath[MAXN];
    for(int i = 1; i <= n; ++i)Curhead[i] = head[i];
    u = src;
    while(dist[src] < n)
    {
        if(u == des)     // find an augmenting path
        {
            int augflow = INF;
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
                augflow = min(augflow, edge[Curhead[i]].cap);
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
            {
                edge[Curhead[i]].cap -= augflow;
                edge[edge[Curhead[i]].rev].cap += augflow;
                edge[Curhead[i]].flow += augflow;
                edge[edge[Curhead[i]].rev].flow -= augflow;
            }
            totalflow += augflow;
            u = src;
        }
        int i;
        for(i = Curhead[u]; i != -1; i = edge[i].next)
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;
        if(i != -1)     // find an admissible arc, then Advance
        {
            Curhead[u] = i;
            revpath[edge[i].ver] = edge[i].rev;
            u = edge[i].ver;
        }
        else        // no admissible arc, then relabel this vertex
        {
            if(0 == (--numbs[dist[u]]))break;    // GAP cut, Important!
            Curhead[u] = head[u];
            int mindist = n;
            for(int j = head[u]; j != -1; j = edge[j].next)
                if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);
            dist[u] = mindist + 1;
            ++numbs[dist[u]];
            if(u != src)
                u = edge[revpath[u]].ver;    // Backtrack
        }
    }
    return totalflow;
}
int in()
{
    int flag = 1;
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' || ch == '\n');
    if(ch == '-') flag = -1;
    else
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return flag * a;
}
int main()
{
    int p, t;
    while(scanf("%d%d%d", &n, &p, &t) != EOF)
    {
        int mi = INF, mx = 0;
        for(int i = 1; i <= p; i++)
        {
            xx[i] = in();
            yy[i] = in();
            cc[i] = in();
            mi = min(mi, cc[i]);
            mx = max(mx, cc[i]);
        }
        int low = mi, high = mx, ans = mx;
        while(low <= high)
        {
            int mid = (low + high) >> 1;
            e = 0;
            memset(head, -1, sizeof(head));
            memset(dist, 0, sizeof(dist));
            memset(numbs, 0, sizeof(numbs));
            src = 1;
            des = n;
            for(int i = 1; i <= p; i++)
            {
                if(cc[i] <= mid)
                {
                    add(xx[i], yy[i], 1);
                    add(yy[i], xx[i], 1);
                }
            }
            rev_BFS();
            int tmp = maxflow();
            //printf("ss %d\n", tmp);
            if(tmp >= t) { ans = min(ans, mid); high = mid - 1;}
            else low = mid + 1;
        }
        printf("%d\n", ans);

    }
    return 0;
}