2016级算法第五次上机-F.ModricWang的水系法术

1066 ModricWang的水系法术

思路

比较典型的最大流问题，需要注意的是，题目已经暗示(明示)了这里的边是双向的，在建图的时候需要加上反向边的容量值。

解决最大流问题的基本思路就是不断在残量网络上找增广路径，这里可以参考一下我院远古学长Song Renfei对于ISAP算法的讲解：ISAP

时间复杂度(O(V^2 sqrt E))

代码

#include <iostream>
#include <cstring>

using std::ios_base;
using std::cin;
using std::cout;

const int MAXN = 1100;
int maze[MAXN][MAXN];
int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
int sap(int start, int end, int nodenum) {
	memset(cur, 0, sizeof(cur));
	memset(dis, 0, sizeof(dis));
	memset(gap, 0, sizeof(gap));
	int u = pre[start] = start, maxflow = 0, aug = -1;
	gap[0] = nodenum;
	while (dis[start] < nodenum) {
		loop:
		for (int v = cur[u]; v < nodenum; v++)
			if (maze[u][v] && dis[u]==dis[v] + 1) {
				if (aug==-1 || aug > maze[u][v])aug = maze[u][v];
				pre[v] = u;
				u = cur[u] = v;
				if (v==end) {
					maxflow += aug;
					for (u = pre[u]; v!=start; v = u, u = pre[u]) {
						maze[u][v] -= aug;
						maze[v][u] += aug;
					}
					aug = -1;
				}
				goto loop;
			}
		int mindis = nodenum - 1;
		for (int v = 0; v < nodenum; v++)
			if (maze[u][v] && mindis > dis[v]) {
				cur[u] = v;
				mindis = dis[v];
			}
		if ((--gap[dis[u]])==0)break;
		gap[dis[u] = mindis + 1]++;
		u = pre[u];
	}
	return maxflow;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n, m;
	cin >> n >> m;
	memset(maze, 0, sizeof(maze));
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		maze[a][b] = maze[b][a] = c;
	}
	int ans = sap(1, n, n + 1);
	cout << ans << "
";
}