LA 6474 Drop Zone 解题报告

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\n", SAP (SS, ST, n * m + 1) );
    }
    return 0;
}