uva 10603 倒水问题

刘汝佳那本书上的代码有些问题，判断当前三个瓶子的水量状态是否被访问过后，还要判断当前状态的使用水量是否比之前的要少。如果少那么需要更新到达该状态的水量。

但是奇怪的是，缺少这一判断的代码依旧能够通过测试，目前不知道是由于缺少相应测试还是个人想法错了

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string>#include <string.h>#include <algorithm>#include <queue>using namespace std;int cap[3], d;struct Node{int val[3];int dist;Node(int a = 0, int b = 0, int c = 0){val[0] = a;val[1] = b;val[2] = c;}};bool operator < (const Node& a, const Node& b){return a.dist > b.dist;}const int maxn = 200 + 10;bool state[maxn][maxn];int res[maxn];void update_dist(const Node & node){for (int i = 0; i < 3; i++){if (res[node.val[i]] < 0 || res[node.val[i]] > node.dist)res[node.val[i]] = node.dist;}}void work(){memset(res, -1, sizeof(res));memset(state, false, sizeof(state));priority_queue<Node> que;Node start(0, 0, cap[2]);start.dist = 0;que.push(start);state[0][0] = true;while (!que.empty()){Node nt = que.top();que.pop();update_dist(nt);for (int i = 0; i < 3; i++){for (int j = 0; j < 3; j++){if (i == j || !nt.val[i] || nt.val[j] == cap[j])continue;int water_amount = min(cap[j], nt.val[i] + nt.val[j]) - nt.val[j];Node ssnt;memcpy(&ssnt, &nt, sizeof(Node));ssnt.val[i] -= water_amount;ssnt.val[j] += water_amount;ssnt.dist += water_amount;if (!state[ssnt.val[0]][ssnt.val[1]]){state[ssnt.val[0]][ssnt.val[1]] = true;que.push(ssnt);}elseupdate_dist(ssnt); // 这个更新存在与否都能够通过过测试， 目前没有想清楚}//for int j}//for int i}//while que.empty()while (d >= 0){if (res[d] >= 0){printf("%d %d\n", res[d], d);return;}d--;}}int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);int kasenum;scanf("%d", &kasenum);while (kasenum--){for (int i = 0; i < 3; i++)scanf("%d", &cap[i]);scanf("%d", &d);work();}//whilereturn 0;}