倒水问题 Fill,UVa 10603

#include"stdafx.h"
// UVa10603 Fill
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

struct Node {
	int v[3], dist;
	bool operator < (const Node& rhs) const {
		return dist > rhs.dist;
	}
};

const int maxn = 200 + 5;
int mark[maxn][maxn], dist[maxn][maxn], cap[3], ans[maxn];

void update_ans(const Node& u) {
	for (int i = 0; i < 3; i++) {
		int d = u.v[i];
		if (ans[d] < 0 || u.dist < ans[d]) ans[d] = u.dist;   //新节点添加后,更新每个值 好像dijkstra
	}
}

void solve(int a, int b, int c, int d) {
	cap[0] = a; cap[1] = b; cap[2] = c;
	memset(ans, -1, sizeof(ans));
	memset(mark, 0, sizeof(mark));
	memset(dist, -1, sizeof(dist));
	priority_queue<Node> q;

	Node start;
	start.dist = 0;
	start.v[0] = 0; start.v[1] = 0; start.v[2] = c;
	q.push(start);

	dist[0][0] = 0;
	while (!q.empty()) {
		Node u = q.top(); q.pop();
		if (mark[u.v[0]][u.v[1]]) continue;
		mark[u.v[0]][u.v[1]] = 1;
		update_ans(u);
		if (ans[d] >= 0) break;
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++) if (i != j) {
				if (u.v[i] == 0 || u.v[j] == cap[j]) continue;   //没水或被倒的杯子中水是满的
				int amount = min(cap[j], u.v[i] + u.v[j]) - u.v[j];  //i向j倒水,如果到过后总水量(此时j的水量+倒入的水量)超过i的最大水量(倒多了),既最多只能倒cap[j]的水量,然后剪去初始水量,既到了多少水
				Node u2;
				memcpy(&u2, &u, sizeof(u));  //增加新节点
				u2.dist = u.dist + amount;
				u2.v[i] -= amount;
				u2.v[j] += amount;
				int& D = dist[u2.v[0]][u2.v[1]];  //用倒过水后,被子1和2的水量做标记,确定一个状态D,类似hash
				if (D < 0 || u2.dist < D) {        //如果是第一次肯定是-1,或这个状态以前存在过,但状态D下,此时的dist比以前的dist小,既总水量小
					D = u2.dist;  //更新dist
					q.push(u2);  //添加节点
				}
			}
	}
	while (d >= 0) {
		if (ans[d] >= 0) {
			printf("%d %d\n", ans[d], d);
			return;
		}
		d--;
	}
}

int main() {
	int T, a, b, c, d;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d%d%d", &a, &b, &c, &d);
		solve(a, b, c, d);
	}
	return 0;
}