Gym 100203I I WIN（最大流）

題意：有一個由字母w，i，n組成的矩陣，問這個矩陣中最多能找到多少個沒有覆寫的win（可以是直線型或者L型）。

思路：最大流。将源點與所有的w連一條容量為1的邊，将所有n與彙點連一條容量為1的邊，将每個字母i拆成兩個點之間連一條容量為1的邊，然後就是根據原圖建立新邊，答案就是最大流。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
//最大流
//Dinic算法求最大流，一般效率很高
//s源點，t彙點，g，e全局變量，表示存邊的鄰接表
const int INF = 1000000000;
const int MAXN = 4000, MAXM = 500000;//最大的點數和邊數
const int DX[] = {-1, 0, 1, 0};
const int DY[] = {0, -1, 0, 1};
struct Edge
{
    int to, next, cap;
} edge[MAXM*2];
int n, s, t;
int head[MAXN], tot;

void addedge(int u, int v, int w)
{
    edge[tot].to = v;
    edge[tot].cap = w;
    edge[tot].next = head[u];
    head[u] = tot++;
    edge[tot].to = u;
    edge[tot].cap = 0;
    edge[tot].next = head[v];
    head[v] = tot++;
}

void init()
{
    memset(head, -1, sizeof(head));
    tot = 0;
}

queue<int> q;
bool vis[MAXN];
int dist[MAXN];
void bfs()
{
    memset(dist, 0, sizeof(dist));
    while(!q.empty()) q.pop();
    vis[s] = 1;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i!=-1; i=edge[i].next)
        {
            if(!edge[i].cap || vis[edge[i].to]) continue;
            q.push(edge[i].to);
            vis[edge[i].to] = 1;
            dist[edge[i].to] = dist[u] + 1;
        }
    }
}

int dfs(int u, int a)
{
    if(u==t) return a;
    int ret = 0;
    for(int i = head[u]; a&&i!=-1; i = edge[i].next)
    {
        if(edge[i].cap && dist[u]+1==dist[edge[i].to])
        {
            int tmp = dfs(edge[i].to, min(a, edge[i].cap));
            edge[i].cap -= tmp;
            edge[i^1].cap += tmp;
            a -= tmp;
            ret += tmp;
        }
    }
    return ret;
}

int maxflow()
{
    int ret = 0;
    while(true)
    {
        memset(vis, 0, sizeof(vis));
        bfs();
        if(!vis[t]) return ret;
        ret += dfs(s, INF);
    }
}
char G[50][50];
int id[50][50];
int main()
{
    //freopen("input.txt", "r", stdin);
	//freopen("output2.txt", "w", stdout);
	int row, col;
	while(cin >> row >> col)
    {
        init();
        int cntW = 0, cntI = 0, cntN = 0;
        for(int i = 1; i <= row; i++)
        {
            scanf("%s", G[i]+1);
            for(int j = 1; j <= col; j++)
            {
                if(G[i][j] == 'W') cntW++;
                else if(G[i][j] == 'I') cntI++;
                else cntN++;
            }
        }
        int tmp1 = 0, tmp2 = 0, tmp3 = 0;
        for(int i = 1; i <= row; i++)
            for(int j = 1; j <= col; j++)
            {
                if(G[i][j] == 'W') id[i][j] = ++tmp1;
                else if(G[i][j] == 'I') id[i][j] = ++tmp2;
                else id[i][j] = ++tmp3;
            }
        for(int i = 1; i <= row; i++)
        {
            for(int j = 1; j <= col; j++)
            {
                if(G[i][j] == 'W')
                {
                    for(int k = 0; k < 4; k++)
                    {
                        int nx = i + DX[k], ny = j + DY[k];
                        if(nx<1 || nx>row || ny<1 || ny>col) continue;
                        if(G[nx][ny] == 'I') addedge(1+id[i][j], 1+cntW+id[nx][ny], 1);
                    }
                }
                else if(G[i][j] == 'N')
                {
                    for(int k = 0; k < 4; k++)
                    {
                        int nx = i + DX[k], ny = j + DY[k];
                        if(nx<1 || nx>row || ny<1 || ny>col) continue;
                        if(G[nx][ny] == 'I') addedge(1+cntW+cntI+id[nx][ny], 1+cntW+cntI+cntI+id[i][j], 1);
                    }
                }
            }
        }
        n = cntW + 2*cntI + cntN + 2;
        s = 1, t = n;
        for(int i = 1; i <= cntI; i++)
            addedge(1+cntW+i, 1+cntW+cntI+i, 1);
        for(int i = 1; i <= cntW; i++)
            addedge(1, i+1, 1);
        for(int i = 1; i <= cntN; i++)
            addedge(1+cntW+cntI+cntI+i, n, 1);
        int ans = maxflow();
        cout << ans << endl;
	}
    return 0;
}