[HEOI2016/TJOI2016]游戏

嘟嘟嘟


这题看数据范围大概能猜出来是网络流。
看到网格图，就会想到这么两种解决方法：黑白染色或每一行每一列看成一个点。
而不管用哪种方法，目的都是建立二分图，把冲突用连边表示出来，而同一侧的点之间没有冲突。


对于这道题，黑白染色肯定gg，但是第二种方法也不是很好，毕竟同一行可能放多个炸弹的。换句话说，这种方法只能解决图中没有石头的情况。
那有石头怎么办？这时候把每一行（列）再拆开一些，把没有‘#‘的一段看成一个点，然后连边即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 55;
const int maxN = 5e3 + 5;
const int maxe = 1e6 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) last = ch, ch = getchar();
    while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

char a[maxn][maxn];
int n, m, cnt = 1, t;
int l[maxn][maxn], r[maxn][maxn];
bool vis[maxN];
struct Edge
{
    int nxt, to, cap, flow;
}e[maxe];
int head[maxN], ecnt = -1;
In void addEdge(int x, int y, int w)
{
    e[++ecnt] = (Edge){head[x], y, w, 0};
    head[x] = ecnt;
    e[++ecnt] = (Edge){head[y], x, 0, 0};
    head[y] = ecnt;
}

int dis[maxN];
In bool bfs()
{
    Mem(dis, 0); dis[0] = 1;
    queue<int> q; q.push(0);
    while(!q.empty())
    {
        int now = q.front(); q.pop();
        for(int i = head[now], v; i != -1; i = e[i].nxt)
        {
            if(!dis[v = e[i].to] && e[i].cap > e[i].flow)
                dis[v] = dis[now] + 1, q.push(v);
        }
    }
    return dis[t];
}
int cur[maxN];
In int dfs(int now, int res)
{
    if(now == t || res == 0) return res;
    int flow = 0, f;
    for(int& i = cur[now], v; i != -1; i = e[i].nxt)
    {
        v = e[i].to;
        if(dis[v] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
        {
            e[i].flow += f; e[i ^ 1].flow -= f;
            flow += f; res -= f;
            if(res == 0) break;
        }
    }
    return flow;
}
In int maxflow()
{
    int flow = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        flow += dfs(0, INF);
    }
    return flow;
}

int main()
{
    Mem(head, -1);
    n = read(), m = read();
    for(int i = 1; i <= n; ++i) scanf("%s", a[i] + 1);
    for(int i = 1; i <= n; ++i, ++cnt)
        for(int j = 1; j <= m; ++j)
            if(a[i][j] == '#') ++cnt;
            else l[i][j] = cnt;
    for(int j = 1; j <= m; ++j, ++cnt)
        for(int i = 1; i <= n; ++i)
            if(a[i][j] == '#') ++cnt;
            else r[i][j] = cnt;
    t = cnt;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
            if(a[i][j] == '*')
            {
                addEdge(l[i][j], r[i][j], 1);
                if(!vis[l[i][j]]) vis[l[i][j]] = 1, addEdge(0, l[i][j], 1); 
            } 
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) if(a[i][j] == '*' && !vis[r[i][j]])
        {
            vis[r[i][j]] = 1;
            addEdge(r[i][j], t, 1);
        }
    write(maxflow()), enter;
    return 0;
}

[HEOI2016/TJOI2016]游戏

原文：https://www.cnblogs.com/mrclr/p/10816783.html