Gym 100203I I WIN（最大流）

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题意：有一个由字母w，i，n组成的矩阵，问这个矩阵中最多能找到多少个没有覆盖的win（可以是直线型或者L型）。

思路：最大流。将源点与所有的w连一条容量为1的边，将所有n与汇点连一条容量为1的边，将每个字母i拆成两个点之间连一条容量为1的边，然后就是根据原图建立新边，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源点，t汇点，g，e全局变量，表示存边的邻接表const int INF = 1000000000;const int MAXN = 4000, MAXM = 500000;//最大的点数和边数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;}

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