POJ 2391.Ombrophobic Bovines 解题报告

实际上是求最短的避雨时间。

首先将每个点拆成两个，一个连接源点，一个连接汇点，连接源点的点的容量为当前单的奶牛数，连接汇点的点为能容纳的奶牛数。

floyd求任意两点互相到达的最短时间，二分最长时间，最大流判断是否可行。

注意路径时间会超过int

/*
      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 = 500;
long long G[INF][INF];
struct node {
    int v, c, next;
} edge[INF*INF * 4];
long long  pHead[INF*INF], SS, ST, nCnt;
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) {
    int numh[INF], h[INF], curEdge[INF], pre[INF];
    int cur_flow, flow_ans = 0, u, neck, i, tmp;
    ms (h, 0); ms (numh, 0); ms (pre, -1);
    for (i = 0; i <= N; i++) curEdge[i] = pHead[i];
    numh[0] = N;
    u = pStart;
    while (h[pStart] <= N) {
        if (u == pEnd) {
            cur_flow = 1e9;
            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;
            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;
            u = neck;
        }
        for ( i = curEdge[u]; i != 0; i = edge[i].next)
            if (edge[i].c && h[u] == h[edge[i].v] + 1)     break;
        if (i != 0) {
            curEdge[u] = i, pre[edge[i].v] = u;
            u = edge[i].v;
        }
        else {
            if (0 == --numh[h[u]]) continue;
            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;
}
long long m, n, x, y,sum,c;
int in[INF], out[INF];
bool check (long long tem) {
    nCnt = 1;
    SS = 2 * n + 1, ST = 2 * n + 2;
    memset (pHead, 0, sizeof pHead);
    for (int i = 1; i <= n; i++) {
        if(out[i]) addEdge (SS, i, out[i]);
        for (int j =1; j <= n; j++)
            if (G[i][j] <= tem&&G[i][j]!=-1)
                addEdge (i, j+n, 5000);
    }
    for (int i = 1; i <= n; i++)
              if(in[i])addEdge (i + n, ST, in[i]);
    int ans = SAP (SS, ST, ST);
    if (ans == sum) 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
    */
    ms (G, -1);
    cin>>n>>m;
    for (int i = 1; i <= n; i++) {
        cin>>out[i]>>in[i];
        sum += out[i];
    }
    long long l = 0x7fffffffffffffff, r = 0;
    for (int i = 1; i <= n; i++) G[i][i] = 0;
    for (int i = 1; i <= m; i++) {
        cin>>x>>y>>c;
        if(G[x][y]>0) G[x][y] = G[y][x] = min(c,G[x][y]);
        else
                     G[x][y]=G[y][x]=c;
        l = min (l, c), r = max (r, c);
    }
    for (int  t = 1; t <= n; t++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++) {
                if (G[i][t] == -1 || G[t][j] == -1) continue;
                if (G[i][j] == -1 || G[i][j] > G[i][t] + G[t][j])
                    G[i][j] = G[i][t] + G[t][j], l = min (l, G[i][j]), r = max (r, G[i][j]);
            }
    long long last = -1,mid;
    while (l <= r) {
        mid = (l + r) >> 1;
        if (check (mid) ) {
            last = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    cout<<last;
    return 0;
}