UVA 1515Pool construction

题目大意：给出一个字符矩阵，其中有草地‘＃’和洞‘.’可以将草改成洞花费d，也可以把洞改成草花费f，最后需要在草和洞之间修围栏，费用b，要求用最少的费用，要求第一行／列和最后一行／列必须为草

解题思路：先判断边界，如果为‘.’可以先填为‘＃’并增加费用，之后构建网络流，做最小割问题，将源点s与所有‘＃’相连，容量为d，代表要必须要割掉这条边才能将‘＃’变为‘.’同理，将所有‘.’与t相连，容量f，然后将所有相邻格子连俩条方向不同的线，容量为b，表示若俩格子一个为‘＃’一个为‘.’时，需要切割掉其中一条，最后求用最大流算法求出最小割就是答案

#include <iostream>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define maxn 4000int x, y, d, f, b;int result;int s, t;char ip[maxn][maxn];int dir[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};struct Edge {    int from, to, cap, flow;    Edge(int from, int to, int cap, int flow):from(from), to(to), cap(cap), flow(flow){}};vector<Edge> edges;vector<int> G[maxn];void addEdge(int from, int to, int cap) {    edges.push_back(Edge(from, to, cap, 0));    edges.push_back(Edge(to, from, 0, 0));    int i = edges.size();    G[from].push_back(i-2);    G[to].push_back(i-1);}int EK() {    int re = 0;    int flo[maxn];    int pre[maxn];    while(1) {        memset(flo, 0, sizeof(flo));        flo[s] = INF;        queue<int> que;        que.push(s);        while(!que.empty()) {            int po = que.front();            que.pop();            for(int i = 0; i < G[po].size(); i++) {                Edge& ed = edges[G[po][i]];                if(!flo[ed.to] && ed.cap > ed.flow ) {                    flo[ed.to] = min(ed.cap - ed.flow, flo[po]);                    pre[ed.to] = G[po][i];                    que.push(ed.to);                }            }            if(flo[t])                break;        }        if(!flo[t])            break;        for(int i = t; i != s; i = edges[pre[i]].from) {            edges[pre[i]].flow += flo[t];            edges[pre[i]^1].flow -= flo[t];        }        re += flo[t];    }    return re;}void init() {    s = 0;    t = x * y + 1;    result = 0;    edges.clear();    for(int i = 0; i <= x*y+2; i++)        G[i].clear();    for(int i = 1; i <=  x; i++) {        for(int j = 1; j <= y; j++) {            if(i != x && j != 1 && i != 1 && j != y) {                if(ip[i][j] == '#') {                    addEdge(s, (i-1)*y+j, d);                }                else {                    addEdge((i-1)*y+j, t, f);                }            }            else {                if(ip[i][j] == '.') {                    result += f;                }                addEdge(s, (i-1)*y+j, INF);            }            for(int k = 0; k < 4; k++) {                int ci = i + dir[k][0];                int cj = j + dir[k][1];                if(ci <= 0 || ci > x || cj <= 0 || cj > y)                    continue;                addEdge((i-1)*y+j, (ci-1)*y+cj, b);            }        }    }}int main() {    int n;    cin >> n;    while(n--) {        cin >> y >> x;        cin >> d >> f >> b;        getchar();        for(int i = 1; i <= x; i++) {            cin >> ip[i]+1;        }        init();        result += EK();        cout << result << endl;    }    return 0;}