Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Superior Island is a very picturesque island and only bicycles are allowed on the island. Therefore, there are many one-way bicycle roads connecting the dierent best photo-shooting spots on the island. To help the visitors plan their trip to the island, the tourism commission wants to designate r different bicycle routes that go through some of the best photo-shooting spots on the island. Given a map of all the bicycle roads on the island and a list of the best photo-shooting spots to be included on each of the three planned routes (non-listed spots must not be included in the route), please write a program to plan each of the r routes so that the distance on each route is minimal. Note that each best photo-shooting spot may only appear at most once on the route.

Input 

There are two parts to the input. The first part of input gives the information of the bicycle roads on the island. The first line contains two integer n and r , n100 and r10 , indicating that there are n best photo-shooting spots on the island and there are r routes to be planned. The next n lines (line 2 through line n + 1 ) contains n×n integers (n lines with n integers on each line), where the j -th integer on line i denotes the distance from best photo-shooting spot i - 1 to best photo-shooting spot j ; the distances are all between 0 and 10, where 0 indicates that there is no one-way road going from best photo-shooting spot i - 1 to spot j .

The second part of input has r lines, denoting the r sightseeing routes to be planed. Each line lists the best photo-shooting stops to be included in that route. The integers on each line denote the recommended photo-shooting stops on that particular sightseeing route. The first integer on the line is the starting point of the route and the last integer is the last stop on the route. However, the stops in between can be visited in any order.

Output 

Output r integers on r lines (one integer per line) indicating the distance of each of the r planned routes. If a route is not possible, output `0'.

Sample Input 

6 3
0 1 2 0 1 1
1 0 1 1 1 0
0 2 0 1 3 0
4 3 1 0 0 0
0 0 1 1 0 0
1 0 0 0 0 0
1 3 5
6 3 2 5
6 1 2 3 4 5

Sample Output 

5 
0 
7

---

每定起點與終點，圖中經過的點任你排序，求總路徑長最小。
n <= 100，因此不太可能狀態壓縮，但是測資中路徑上的點不超過 20，
但為了保險起見，使用 A*，並且將所有狀態點壓成 4 個 int 的 bitmask，
H function 則採用使用剩餘點的單源最短路徑到終點。 

#include <stdio.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <queue>
#include <string.h>
using namespace std;
struct E {
    int mark[5], used;
    int last;
    int g, h;//f = g + h
    E() {
        memset(mark, 0, sizeof(mark));
        last = g = h = 0;
    }
    int GET(int x) {
        return mark[x>>5]>>(x&31)&1;
    }
    void SET(int x) {
        mark[(x)>>5] |= 1<<(x&31);
    }
    bool operator<(const E &a) const {
        if(g+h != a.g+a.h)
            return g+h > a.g+a.h;
        if(g != a.g)
            return g < a.g;
        if(used != a.used)
            return used < a.used;
        return memcmp(mark, a.mark, sizeof(mark)) > 0;
    }
};
int g[105][105], f[105][105];
int rcnt = 0, route[105];
int h(E tn) {//single shortest path to END node.
    queue<int> Q;
    static int dist[105], inq[105], i, tnode;
    for(i = 0; i < rcnt; i++)
        dist[i] = 0xfffffff, inq[i] = 0;
    Q.push(tn.last);
    dist[tn.last] = 0;
    while(!Q.empty()) {
        tnode = Q.front(), Q.pop();
        inq[tnode] = 0;
        for(i = 0; i < rcnt; i++) {
            if(f[tnode][i] && tn.GET(i) == 0) {
                if(dist[i] > dist[tnode] + f[tnode][i]) {
                    dist[i] = dist[tnode] + f[tnode][i];
                    if(inq[i] == 0)
                        inq[i] = 1, Q.push(i);
                }
            }
        }
    }
    return dist[rcnt-1] != 0xfffffff ? dist[rcnt-1] : -1;
}
int astar(int route[], int rcnt, int n) {
    int i, j, k;
    int st = route[0], ed = route[n-1];
    for(i = 0; i < rcnt; i++)
        for(j = 0; j < rcnt; j++)
            f[i][j] = g[route[i]][route[j]];
    priority_queue<E, vector<E> > pQ;
    E tn;
    tn.SET(0);//START node used.
    tn.last = 0;
    tn.g = 0, tn.h = h(tn);
    tn.used = 1;
    if(tn.h == -1)  return -1;
    pQ.push(tn);
    while(!pQ.empty()) {
        tn = pQ.top(), pQ.pop();
        if(tn.used == rcnt) {
            return tn.g;
        }
        //printf("%d ------\n", tn.used);
        for(i = 0; i < rcnt; i++) {
            if(tn.GET(i) == 0 && f[tn.last][i]) {
                if(i == rcnt-1 && tn.used+1 != rcnt)
                    continue;
                E tm = tn;
                tm.SET(i), tm.last = i;
                tm.g = tn.g + f[tn.last][i];
                tm.h = h(tm);
                tm.used++;
                if(tm.h < 0)    continue;
                pQ.push(tm);
            }
        }
    }
    return -1;
}
int main() {
    int n, m;
    int i, j, k;
    char s[1024];
    while(scanf("%d %d", &n, &m) == 2) {
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                scanf("%d", &g[i][j]);
        while(getchar() != '\n');
        while(m--) {
            gets(s);
            stringstream sin(s);
            rcnt = 0;
            while(sin >> route[rcnt])
                rcnt++;
            int ret = astar(route, rcnt, n);
            printf("%d\n", ret < 0 ? 0 : ret);
        }
    }
    return 0;
}