• LA 6474 Drop Zone (最小割)


    题目链接

      要添最少的挡板使所有的'D'不存在到达网格外的路径.

          以每个格子向四个方向中可以到达的格子连容量为1的边, 从源点向所有'D' 连容量为4的边,网格外的点向汇点连一条容量为4的边.

          答案就是这个容量网络的最小割,即最大流.

    /*
          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>
    #include <utility>
    #include <vector>
    #define ms(a,b) memset(a,b,sizeof a)
    using namespace std;
    const int INF = 6111;
    struct node {
        int v, c, next;
    } edge[100000];
    int  pHead[100000], SS, ST, nCnt;
    int n, m;
    int g[200][200];
    int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
    //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;
    }
    inline int SAP (int pStart, int pEnd, int N) {
        //层次点的数量  点的层次   点的允许弧     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;
    }
    inline void build() {
        char ch;
        scanf ("%d %d", &n, &m);
        ms (g, -1), ms (pHead, 0), nCnt = 1;
        for (int i = 1; i <= n; i++) {
            getchar();
            for (int j = 1; j <= m; j++) {
                ch = getchar();
                if (ch == '.')  g[i][j] = 0;
                if (ch == 'D') g[i][j] = 1;
            }
        }
        n += 2, m += 2;
        SS = n * m, ST = SS + 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int u = i * m + j;
                if (i == 0 || i == n - 1 || j == 0 || j == m - 1) {
                    addEdge (u, ST, 4);
                    continue;
                }
                if (g[i][j] == 0)  {
                    for (int k = 0; k < 4; k++) {
                        int x = i + dx[k], y = j + dy[k];
                        int v = m * x + y;
                        if (g[x][y] != 1)    addEdge (u, v, 1);
                    }
                }
                if (g[i][j] == 1) {
                    addEdge (SS, u, 4);
                    for (int k = 0; k < 4; k++) {
                        int x = i + dx[k], y = j + dy[k];
                        int v = m * x + y;
                        if (g[x][y] != 1 )     addEdge (u, v, 1);
                    }
                }
            }
        }
    }
    int cs;
    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", &cs);
        while (cs--) {
            build();
            printf ("%d
    ", SAP (SS, ST, n * m + 1) );
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/keam37/p/4293602.html
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