POJ 2112.Optimal Milking 解题报告

/*
      最大流SAP
      邻接表
      思路：基本源于FF方法，给每个顶点设定层次标号，和允许弧。
      优化:
      1、当前弧优化（重要）。
      1、每找到以条增广路回退到断点（常数优化）。
      2、层次出现断层，无法得到新流（重要）。
      时间复杂度（m*n^2）
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
const int INF = 300;
int G[INF][INF];
struct node {
    int v, c, next;
} edge[INF*INF*4];
int  pHead[INF*INF], SS, ST, nCnt;
//同时添加弧和反向边, 反向边初始容量为0
void addEdge (int u, int v, int c) {
    edge[++nCnt].v = v; edge[nCnt].c = c, edge[nCnt].next = pHead[u]; pHead[u] = nCnt;
    edge[++nCnt].v = u; edge[nCnt].c = 0, edge[nCnt].next = pHead[v]; pHead[v] = nCnt;
}
int SAP (int pStart, int pEnd, int N) {
    //层次点的数量  点的层次   点if(G[i][j]<l) l=G[i][j];的允许弧     当前走过边的栈
    int numh[INF], h[INF], curEdge[INF], pre[INF];
    //当前找到的流, 累计的流量, 当前点, 断点, 中间变量
    int cur_flow, flow_ans = 0, u, neck, i, tmp;
    //清空层次数组,
    ms (h, 0); ms (numh, 0); ms (pre, -1);
    //将允许弧设为邻接表的任意if(G[i][j]<l) l=G[i][j];一条边
    for (i = 0; i <= N; i++) curEdge[i] = pHead[i];
    numh[0] = N;//初始全部点的层次为0
    u = pStart;//从源点开始
    //如果从源点能找到增广路
    while (h[pStart] <= N) {
        //找到增广路
        if (u == pEnd) {
            cur_flow = 1e9;
            //找到当前增广路中的最大流量, 更新断点
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v)
                if (cur_flow > edge[curEdge[i]].c) neck = i, cur_flow = edge[curEdge[i]].c;
            //增加反向边的容量
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v) {
                tmp = curEdge[i];
                edge[tmp].c -= cur_flow, edge[tmp ^ 1].c += cur_flow;
            }
            flow_ans += cur_flow;//累计流量
            u = neck;//从断点开始找新的增广路
        }
        //找到一条允许弧
        for ( i = curEdge[u]; i != 0; i = edge[i].next)
            if (edge[i].c && h[u] == h[edge[i].v] + 1)     break;
        //继续DFS
        if (i != 0) {
            curEdge[u] = i, pre[edge[i].v] = u;
            u = edge[i].v;
        }
        //当前起点没有允许弧,从u找不到增广路
        else {
            //u所在的层次点减少一,且如果没有与当前点一个层次的点, 退出.
            if (0 == --numh[h[u]]) continue;
            //有与u相同层次的点, 更新u的层次 ,回到上一个点
            curEdge[u] = pHead[u];
            for (tmp = N, i = pHead[u]; i != 0; i = edge[i].next)
                if (edge[i].c)  tmp = min (tmp, h[edge[i].v]);
            h[u] = tmp + 1;
            ++numh[h[u]];
            if (u != pStart) u = pre[u];
        }
    }
    return flow_ans;
}
int k, c, m, n;
bool check (int tem) {
    nCnt = 1;
    SS = n + 1, ST = n + 2;
    memset (pHead, 0, sizeof pHead);
    for (int i = 1; i <= k; i++) {
        addEdge (i, ST, m);
        for (int j = k + 1; j <= k + c; j++)
            if (G[j][i] <= tem)
                addEdge (j, i, 1);
    }
    for (int i = k + 1; i <= k + c; i++) addEdge (SS, i, 1);
    int ans = SAP (SS, ST, ST);
    if (ans == c) return 1;
    return 0;
}
int main() {
    /*
           建图,前向星存边，表头在pHead[],边计数 nCnt.
           SS,ST分别为源点和汇点
    */
    scanf ("%d %d %d", &k, &c, &m);
    n = k + c;
    int l = 0, r = 10000;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            scanf ("%d", &G[i][j]);
            if (G[i][j]==0)
                G[i][j] = 0x3f3f3f;
        }
    for (int  t = 1; t <= n; t++) {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (G[i][j] > G[i][t] + G[t][j]) G[i][j] = G[i][t] + G[t][j];
    }
    int last = -1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check (mid) ) {
            last = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    printf ("%d", last);
    return 0;
}