UVA10603 FILL（寬搜）

There are three jugs with a volume of a, b and c liters. (a, b, and c are positive integers not greater than 200). The first and the second jug are initially empty, while the third is completely filled with water. It is allowed to pour water from one jug into another until either the first one is empty or the second one is full. This operation can be performed zero, one or more times.

You are to write a program that computes the least total amount of water that needs to be poured; so that at least one of the jugs contains exactly d liters of water (d is a positive integer not greater than 200). If it is not possible to measure d liters this way your program should find a smaller amount of water d′ < d which is closest to d and for which d′ liters could be produced. When d′ is found, your program should compute the least total amount of poured water needed to produce d′ liters in at least one of the jugs.

Input

The first line of input contains the number of test cases. In the next T lines, T test cases follow. Each test case is given in one line of input containing four space separated integers — a, b, c and d.

Output

The output consists of two integers separated by a single space. The first integer equals the least total amount (the sum of all waters you pour from one jug to another) of poured water. The second integer equals d, if d liters of water could be produced by such transformations, or equals the closest smaller value d′ that your program has found.

Sample Input

2 2 3 4 2 96 97 199 62

Sample Output

2 2 9859 62

首先是狀态數量的判斷：a有（a+1）種，b有（b+1）種，而且a，b<200，是以最多201*201種狀态，用搜尋是可以的。

為了找到最少的水的量（不是步數），定義一個節點包含目前水的數量，用優先隊列來實作最短路。

書上說和Dijkstra很像，還真的是這樣，圖的算法和這種搜尋的算法有着某種神秘的聯系。

是否可以把倒水看成在圖中走一步，該邊的權值就是水量？貪心的往下走？

#include <bits/stdc++.h>
using namespace std;
#define N 205

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

int ans[N * N], cap[3];
bool vis[N][N];

void solve(int a, int b, int c, int d)
{
    memset(vis, 0, sizeof(vis));
    memset(ans, -1, sizeof(ans));

    priority_queue< Node, vector<Node> > que;

    Node s;
    cap[0] = a;cap[1] = b;cap[2] = c;
    s.v[0] = s.v[1] = 0;s.v[2] = c;
    s.dist = 0;
    que.push(s);

    while(!que.empty()){
        Node h = que.top();que.pop();

        for(int i = 0; i < 3; i++){
            if(ans[h.v[i]] < 0 || ans[h.v[i]] > h.dist){
                ans[h.v[i]] = h.dist;
            }
        }

        if(ans[d] >= 0)
            break;

        for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++){
            if(i != j && h.v[i] > 0 && h.v[j] < cap[j]){
                int amount = min(cap[j], h.v[i] + h.v[j]) - h.v[j];

                Node tmp;
                memcpy(&tmp, &h, sizeof(h));
                tmp.dist = h.dist + amount;
                tmp.v[i] -= amount;tmp.v[j] += amount;

                if(!vis[tmp.v[0]][tmp.v[1]]){
                    vis[tmp.v[0]][tmp.v[1]] = true;
                    que.push(tmp);
                }
            }
        }
    }

    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;
}