【BZOJ1324】Exca王者之剑【最小割】

【题目链接】

论文题，见 胡伯涛《最小割模型在信息学竞赛中的应用》

WA了几发，最后发现是dinic模板写错了= =，看来今天不适合做题。

/* Pigonometry */
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 10005, maxm = 100005, maxq = 50000, inf = 0x7fffffff;

int n, m, head[maxn], cur[maxn], cnt, depth[maxn], bg, ed, q[maxq];

struct _edge {
	int v, w, next;
} g[maxm << 1];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v, int w) {
	g[cnt] = (_edge){v, w, head[u]};
	head[u] = cnt++;
}

inline void insert(int u, int v, int w) {
	add(u, v, w); add(v, u, 0);
}

inline bool bfs() {
	for(int i = 0; i <= ed; i++) depth[i] = -1;
	int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;
	while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {
		depth[g[i].v] = depth[u] + 1;
		if(g[i].v == ed) return 1;
		q[t++] = g[i].v;
	}
	return 0;
}

inline int dfs(int x, int flow) {
	if(x == ed) return flow;
	int left = flow;
	for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {
		int tmp = dfs(g[i].v, min(left, g[i].w));
		left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;
		if(g[i].w) cur[x] = i;
		if(!left) return flow;
	}
	if(left == flow) depth[x] = -1;
	return flow - left;
}

int main() {
	n = iread(); m = iread(); bg = 0; ed = n * m + 1;
	for(int i = 0; i <= ed; i++) head[i] = -1; cnt = 0;

	int ans = 0;
	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) {
		int x = iread(); ans += x;
		if((i + j) & 1) {
			insert(bg, (i - 1) * m + j, x);
			if(i != 1) insert((i - 1) * m + j, (i - 2) * m + j, inf);
			if(j != 1) insert((i - 1) * m + j, (i - 1) * m + j - 1, inf);
			if(i != n) insert((i - 1) * m + j, i * m + j, inf);
			if(j != m) insert((i - 1) * m + j, (i - 1) * m + j + 1, inf);
		} else insert((i - 1) * m + j, ed, x);
	}

	while(bfs()) {
		for(int i = 0; i <= ed; i++) cur[i] = head[i];
		ans -= dfs(bg, inf);
	}

	printf("%d\n", ans);
	return 0;
}