题意给出3个杯子，只有最后一个杯子装满水，问至少倒多少水才能使其中一个杯子里的水达到d升。
分析：这题关键在于优先级队列，每次都要取倒水量最小的状态出来。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<queue>
using namespace std;
struct Node {
int v[3], dist;
bool operator<(const Node& u) const{
return dist > u.dist;
}
};
const int maxn = 200 + 5;
int vis[maxn][maxn], cap[3], ans[maxn];
void bfs(int a, int b, int c, int d) {
priority_queue<Node>q;//优先级队列
memset(vis, 0, sizeof(vis));
memset(ans, -1, sizeof(ans));
cap[0] = a, cap[1] = b, cap[2] = c;
Node start;
start.dist = 0;
start.v[0] = 0; start.v[1] = 0; start.v[2] = c;
q.push(start);
vis[0][0] = 1;
while (!q.empty()) {
Node u = q.top(); q.pop();
for (int i = 0; i < 3; i++) {
int d = u.v[i];
if (ans[d]<0 || ans[d]>u.dist)ans[d] = u.dist;//更新每个杯子水的状态信息
}
if (ans[d] >= 0)break;//有目标状态信息，退出
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (u.v[i] == 0 || u.v[j] == cap[j])continue;//杯子满了或者空了
int amount = min(cap[j], u.v[j] + u.v[i]) - u.v[j];//加水量最小
Node u2;
memcpy(&u2, &u, sizeof(u));
u2.dist = u.dist + amount;//u2的倒水量
u2.v[i] -= amount;
u2.v[j] += amount;
if (!vis[u2.v[0]][u2.v[1]]) {
vis[u2.v[0]][u2.v[1]] = 1;
q.push(u2);
}
}
}
}
while (d >= 0) {
if (ans[d] >= 0) {
printf("%d %d\n",ans[d], d);
return;
}
d--;
}
}
int main() {
int kase,a,b,c,d;
cin >> kase;
while (kase-- > 0) {
cin >> a >> b >> c >> d;
bfs(a, b, c, d);
}
//system("pause");
return 0;
}