hdu_1026(最短路)

题意：最快出去的路径

题解：一看就是一个很简单的最短路，用优先队列写个spfa就过了，但是没想到水了个(n*n)^2的算法也过了。。。愿意很简单，因为是个稀疏图

       敲完以后一百度发现就是个简单的bfs+优先队列其实思路是一样的，每次选出不在路径中距离路径最短的边，因为这个更新的时候只能更新其周围的四个点，那个吗。。。懒得打了，。。。记得提醒我有时间打一下。。。。

代码：转化成图后，单源最短路求到n*m-1就可以了，完美

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 111;
#define INF 0x1fffffff
char mp[N][N];
int dist[N*N];
bool p[N*N];
int head[N*N];
struct Edge{
    int to;
    int next;
    int w;
}edge[N*N*10];
int Ecnt;
void init()
{
    Ecnt = 0;
    memset(head,-1,sizeof(head));
}

void printTree(int n, int m){
    for(int i = 0; i < n*m; i++) {
        printf("%d: ", i);
        for(int j = head[i]; j != -1; j = edge[j].next) {
            printf("to = %d w = %d", edge[j].to,edge[j].w);
        }
        puts("");
    }
}

void add(int from, int to, int w){
    edge[Ecnt].to = to;
    edge[Ecnt].w = w;
    edge[Ecnt].next = head[from];
    head[from]=Ecnt++;
}
int go[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
bool in(int x,int y,int n, int m){
    if(x<n&&x>=0&&y<m&&y>=0) return true;
    else return false;
}
int ans[N*N];
int fa[N*N];
int sum[N*N];
void f(int n, int m){
    int t = 0;
    int tm = n*m-1;
    sum[t++] = n*m-1;
    while(fa[tm]!=0){
        sum[t++] = fa[tm];
        tm = fa[tm];
    }
    for(int i = t-1; i >= 0; i--){
         ans[t-i] = sum[i];
    }
}
int Count;
void dijk(int s, int n)
{

    int i , j , k ;
    Count = 0;
    for(i = 0 ;i <= n ;i++)
    {
        p[i] = false;
        dist[i] = INF;
    }
    //p[s] = true;
    dist[s] = 0;

    for( i = 0; i < n ; i++)
    {
        int Min = INF ;
        k = 0 ;
        for( j = 0 ; j < n ; j++)
        {
            if(!p[j]&&dist[j]<Min)
            {
                Min = dist[j];
                k = j;
            }
        }

        if(Min == INF) return ;
        p[k] = true;
        //printf("k = %d
",k);
        for(j = head[k]; j != -1 ; j = edge[j].next)
        {
            Edge e = edge[j];
            if(!p[e.to]&&dist[e.to]>dist[k]+e.w){
                dist[e.to] = dist[k]+e.w;
                fa[e.to] = k;
            }
        }
    }
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i = 0; i < n; i++){
            scanf("%s",&mp[i]);
        }
        int id;
        init();
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                id = i*m+j;
                for(int k= 0; k < 4; k++){
                    int tx = i+go[k][0];
                    int ty = j+go[k][1];
                    int tid = tx*m+ty;
                    if(in(tx,ty,n,m)){
                        if(mp[tx][ty]=='.') add(id,tid,1);
                        else if(mp[tx][ty]=='X');//add(id,tid,INF);
                        else add(id,tid,mp[tx][ty]-'0'+1);
                    }
                }
            }
        }
       // printTree(n,m);
        int cnt = 1;
        int c = (n)*(m);
        dijk(0,c);
        f(n,m);

        int t = 0;
        if(dist[c-1]==INF) {
            printf("God please help our poor hero.
FINISH
");
            continue;
        }
        printf("It takes %d seconds to reach the target position, let me show you the way.
",dist[c-1]);
        for(int i = 1;cnt <= dist[c-1] ; i++){
            int sx,sy,dx,dy;
            sx = ans[t]/(m);
            sy = ans[t++]%(m);
            dx = ans[t]/(m);
            dy = ans[t]%(m);
            printf("%ds:(%d,%d)->(%d,%d)
",cnt++,sx,sy,dx,dy);
            if(mp[dx][dy]!='.') {
                for(int i = 0; i < (mp[dx][dy]-'0');i++)
                    printf("%ds:FIGHT AT (%d,%d)
",cnt++,dx,dy);
            }
        }
        puts("FINISH");
    }
    return 0;
}