HDU 3442 Three Kingdoms BFS + 优先队列

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    已被周赛虐成挂零狗。
    BFS + 优先队列 == 最短路，貌似是可以这样理解的。这样得到的结果即为花费最少的一条路。貌似初级计划里面有一个类似的题。
    因为受过的伤不会再受，所以需要抽象出第三维来表示受过的哪几种伤（状压）。寒假集训的时候还给大一的选手出过这种题。。。竟然还有脸给大一的讲题。。

曼哈顿距离：在欧几里得空间的固定直角坐标系上两点所形成的线段对轴产生的投影的距离总和。

例如在平面上，坐标（x1, y1）的i点与坐标（x2, y2）的j点的曼哈顿距离为：

d(i,j)=|X1-X2|+|Y1-Y2|.

#include <iostream>#include <algorithm>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <cmath>#include <stack>#pragma comment(linker, "/STACK:1024000000");#define EPS (1e-6)#define LL long long#define ULL unsigned long long int#define _LL __int64#define _INF 0x3f3f3f3f#define INF (1<<62)#define Mod 1000000007using namespace std;char Map[51][51];int hurt[51][51][5];bool mark[51][51][33];void Init_Hurt(int n,int m){    int i,j,k,l;    memset(hurt,0,sizeof(hurt));    for(i =1;i <= n; ++i)    {        for(j = 1;j <= m ; ++j)        {            if(Map[i][j] == 'A')            {                for(k = i-2;k <= i+2; ++k)                {                    for(l = j-2;l <= j+2; ++l)                    {                        if(1 <= k && k <= n && 1 <= l && l <= m && abs(k-i)+ abs(l-j) <= 2)                        {                            hurt[k][l][0] = 1;                        }                    }                }            }            else if(Map[i][j] == 'B')            {                for(k = i-3;k <= i+3; ++k)                {                    for(l = j-3;l <= j+3; ++l)                    {                        if(1 <= k && k <= n && 1 <= l && l <= m && abs(k-i)+ abs(l-j) <= 3)                        {                            hurt[k][l][1] = 2;                        }                    }                }            }            else if(Map[i][j] == 'C')            {                hurt[i][j][2] = 3;            }            else if(Map[i][j] == 'D')            {                for(k = i-2;k <= i+2; ++k)                {                    for(l = j-2;l <= j+2; ++l)                    {                        if(1 <= k && k <= n && 1 <= l && l <= m && abs(k-i)+ abs(l-j) <= 2)                        {                            hurt[k][l][3] = 4;                        }                    }                }            }            else if(Map[i][j] == 'E')            {                for(k = i-1;k <= i+1; ++k)                {                    for(l = j-1;l <= j+1; ++l)                    {                        if(1 <= k && k <= n && 1 <= l && l <= m && abs(k-i)+ abs(l-j) <= 1)                        {                            hurt[k][l][4] = 5;                        }                    }                }            }        }    }}struct N{    int x,y,sta,ans;    bool operator < (const N &a) const    {        return a.ans < ans;    }};int jx[] = {-1, 0, 1, 0};int jy[] = { 0, 1, 0,-1};bool Judge(char c){    if(c == '$' || c == 'C' || c == '!' || c == '.')        return true;    return false;}void bfs(int n,int m,int s,int e){    priority_queue<N> q;    N f,t;    f.x = s,f.y = e,f.sta = 0,f.ans = 0;    memset(mark,false,sizeof(mark));    mark[s][e][0] = true;    q.push(f);    while(q.empty() == false)    {        f = q.top();        q.pop();        if(f.x != s || f.y != e)        {            for(int k = 0;k < 5; ++k)            {                if((f.sta&(1<<k)) == 0 && hurt[f.x][f.y][k])                {                    f.ans += hurt[f.x][f.y][k];                    f.sta += (1<<k);                }            }        }        if(Map[f.x][f.y] == '!')        {            printf("%d\n",f.ans);            return ;        }        for(int i = 0;i < 4; ++i)        {            t.ans = f.ans;            t.sta = f.sta;            t.x = f.x + jx[i];            t.y = f.y + jy[i];            if(1 <= t.x && t.x <= n && 1 <= t.y && t.y <= m && mark[t.x][t.y][t.sta] == false && Judge(Map[t.x][t.y]) )            {                q.push(t);                mark[t.x][t.y][t.sta] = true;            }        }    }    printf("-1\n");}void Solve(int n,int m){    int i,j;    for(i = 1; i <= n; ++i)    {        for(j = 1;j <= m; ++j)        {            if(Map[i][j] == '$')            {                bfs(n,m,i,j);                return ;            }        }    }}int main(){    int T,icase = 0;    int i,j,k,n,m;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&m);        for(i = 1;i <= n ; ++i)        {            scanf("%*c%s",Map[i]+1);        }        Init_Hurt(n,m);        printf("Case %d: ",++icase);        Solve(n,m);    }    return 0;}

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