結果

問題 No.1324 Approximate the Matrix
コンテスト
ユーザー 👑 ygussany
提出日時 2020-12-21 23:21:18
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 61 ms / 2,000 ms
コード長 2,884 bytes
コンパイル時間 622 ms
コンパイル使用メモリ 34,048 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-21 13:16:43
合計ジャッジ時間 2,629 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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ファイルパターン 結果
sample AC * 3
other AC * 42
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ソースコード

diff # 

#include <stdio.h>

typedef struct {
	int key, id;
} data;

typedef struct {
	data obj[40001];
	int size;
} min_heap;

void push(data x, min_heap* h)
{
	int i = ++(h->size), j = i >> 1;
	data tmp;
	h->obj[i] = x;
	while (j > 0) {
		if (h->obj[i].key < h->obj[j].key) {
			tmp = h->obj[j];
			h->obj[j] = h->obj[i];
			h->obj[i] = tmp;
			i = j;
			j >>= 1;
		} else break;
	}
}

data pop(min_heap* h)
{
	int i = 1, j = 2;
	data output = h->obj[1], tmp;
	h->obj[1] = h->obj[(h->size)--];
	while (j <= h->size) {
		if (j < h->size && h->obj[j^1].key < h->obj[j].key) j ^= 1;
		if (h->obj[j].key < h->obj[i].key) {
			tmp = h->obj[j];
			h->obj[j] = h->obj[i];
			h->obj[i] = tmp;
			i = j;
			j <<= 1;
		} else break;
	}
	return output;
}

void chmin(int* a, int b)
{
	if (*a > b) *a = b;
}

int main()
{
	int i, j, N, K, A[201], B[201], P[201][201];
	scanf("%d %d", &N, &K);
	for (i = 1; i <= N; i++) scanf("%d", &(A[i]));
	for (i = 1; i <= N; i++) scanf("%d", &(B[i]));
	for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) scanf("%d", &(P[i][j]));
	
	const int sup = 1 << 30;
	int flow[403][403] = {}, cap[403][403] = {}, cost[403][403], pi[403], s = N * 2 + 1, t = N * 2 + 2;
	for (i = 1; i <= N; i++) {
		cap[s][i] = A[i];
		cost[s][i] = 0;
		cap[i+N][t] = B[i];
		cost[i+N][t] = 0;
	}
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			cap[i][j+N] = K;
			cost[i][j+N] = 1 - P[i][j] * 2;
		}
	}
	
	int u, w, n = N * 2 + 2;
	for (i = 1; i <= n; i++) pi[i] = sup;
	for (i = 0, pi[s] = 0; i < 4; i++) {
		for (u = 1; u <= n; u++) {
			if (pi[u] == sup) continue;
			for (w = 1; w <= n; w++) if (cap[u][w] > 0) chmin(&(pi[w]), pi[u] + cost[u][w]);
		}
	}
	
	int k, dist[403], prev[403], flag[403];
	min_heap h;
	data d;
	for (k = 1; k <= K; k++) {
		h.size = 0;
		d.key = 0;
		d.id = s;
		push(d, &h);
		for (u = 1; u <= n; u++) {
			dist[u] = sup;
			prev[u] = -1;
			flag[u] = 0;
		}
		dist[s] = 0;
		prev[s] = s;
		
		while (h.size > 0) {
			d = pop(&h);
			u = d.id;
			if (flag[u] == 1) continue;
			else flag[u] = 1;
			
			for (w = 1; w <= n; w++) {
				if (flow[u][w] < cap[u][w] && dist[w] > dist[u] + cost[u][w] - pi[w] + pi[u]) {
					dist[w] = dist[u] + cost[u][w] - pi[w] + pi[u];
					prev[w] = u;
					d.key = dist[w];
					d.id = w;
					push(d, &h);
				}
			}
		}
		
		for (w = t, u = prev[w]; w != s; w = u, u = prev[w]) {
			if (flow[w][u] > 0) {
				flow[w][u]--;
				cost[w][u] -= 2;
				cost[u][w] += 2;
				if (flow[w][u] == 0) cap[u][w] = 0;
			} else {
				flow[u][w]++;
				if (u <= N * 2 && w <= N * 2) {
					cost[u][w] += 2;
					cost[w][u] = 2 - cost[u][w];
					cap[w][u] = 1;
				}
			}
		}
		for (u = 1; u <= n; u++) if (dist[u] != sup) pi[u] += dist[u];
	}
	
	long long ans = 0;
	for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) ans += (P[i][j] - flow[i][j+N]) * (P[i][j] - flow[i][j+N]);
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}
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