HDU 3338 Kakuro Extension

题目链接

题意：完成如图的游戏，填充数字1-9

思路：网络流的行列模型，把每行每列连续的一段拆分出来建图即可，然后题目有限制一个下限1，所以 每行每列的容量减去相应的数字，然后建图建容量8就好，这样就默认原来容量已经有1了

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 10005;
const int MAXEDGE = 500005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
	}
};

struct Dinic {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool vis[MAXNODE];
	Type d[MAXNODE];
	int cur[MAXNODE];
	vector<int> cut;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}
	void add_Edge(int u, int v, Type cap) {
		edges[m] = Edge(u, v, cap, 0);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bfs() {
		memset(vis, false, sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (!vis[e.v] && e.cap > e.flow) {
					vis[e.v] = true;
					d[e.v] = d[u] + 1;
					Q.push(e.v);
				}
			}
		}
		return vis[t];
	}

	Type dfs(int u, Type a) {
		if (u == t || a == 0) return a;
		Type flow = 0, f;
		for (int &i = cur[u]; i != -1; i = next[i]) {
			Edge& e = edges[i];
			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i^1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}

	Type Maxflow(int s, int t) {
		this->s = s; this->t = t;
		Type flow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v])
				cut.push_back(i);
		}
	}
} gao;

const int N = 105;

int n, m, down[N][N], right[N][N];
int tox[N][N], toy[N][N], xh[N * N], yh[N * N], xr[N * N], yr[N * N];
int xn, yn;

char str[N];

void build(int x, int y) {
	scanf("%s", str);
	if (str[0] == 'X') down[x][y] = -2;
	else if (str[0] == '.') down[x][y] = -1;
	else {
		int num = (str[0] - '0') * 100 + (str[1] - '0') * 10 + str[2] - '0';
		down[x][y] = num;
	}
	if (str[4] == 'X') right[x][y] = -2;
	else if (str[4] == '.') right[x][y] = -1;
	else {
		int num = (str[4] - '0') * 100 + (str[5] - '0') * 10 + str[6] - '0';
		right[x][y] = num;
	}
}

int out[N][N];

int main() {
	while (~scanf("%d%d", &n, &m)) {
		xn = yn = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				build(i, j);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (right[i][j] >= 0)
					xh[++xn] = right[i][j];
				else if (right[i][j] == -1) {
					xh[xn]--;
					tox[i][j] = xn;
					xr[xn] = i;
				}
			}
		}
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if (down[j][i] >= 0)
					yh[++yn] = down[j][i];
				else if (down[j][i] == -1) {
					yh[yn]--;
					toy[j][i] = yn;
					yr[yn] = i;
				}
			}
		}
		gao.init(xn + yn + 2);
		for (int i = 1; i <= xn; i++) gao.add_Edge(0, i, xh[i]);
		for (int i = 1; i <= yn; i++) gao.add_Edge(xn + i, xn + yn + 1, yh[i]);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (right[i][j] == -1)
					gao.add_Edge(tox[i][j], toy[i][j] + xn, 8);
			}
		}
		gao.Maxflow(0, xn + yn + 1);
		for (int i = 0; i < gao.m; i += 2) {
			if (gao.edges[i].u == 0 || gao.edges[i].v == xn + yn + 1) continue;
			out[xr[gao.edges[i].u]][yr[gao.edges[i].v - xn]] = gao.edges[i].flow;
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (right[i][j] == -1) printf("%d", out[i][j] + 1);
				else printf("_");
				printf("%c", j == m - 1 ? '\n' : ' ');
			}
			printf("\n");
		}
	}
	return 0;
}

HDU 3338 Kakuro Extension（网络流）