1026 Ignatius and the Princess I

1026 Ignatius and the Princess I

http://acm.hdu.edu.cn/showproblem.php?pid=1026

 

这道题是经典的地图寻路搜索问题

可以用广度优先搜索来完成此题

AC CODE：

#include <stdio.h>#include <string.h>const int MaxQueue = 1010000;char g[105][105];int v[105][105];int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};int n,m,flagForEnd,second;struct Node{ int x,y; //position int time; int last; // the last step before}state[MaxQueue],temp;void bfs(),init();void output(int t){ if (state[t].last == -1) //如果是第一个点 {// printf("1s:(0,0)->"); return; } output(state[t].last); second++;// if (state[t].last == 0) {printf("(%d,%d)/n",state[t].x,state[t].y);return;}// if (state[t].x == state[state[t].last].x && state[t].y == state[state[t].last].y) //fight// printf("%ds:FIGHT AT (%d,%d)/n",second,state[t].x,state[t].y); if (g[state[t].x][state[t].y]!='.' && g[state[t].x][state[t].y]!='X') { printf("%ds:(%d,%d)->(%d,%d)/n",second++,state[state[t].last].x,state[state[t].last].y,state[t].x,state[t].y); int hp = g[state[t].x][state[t].y] - '0'; int i; for (i=0;i<hp;i++) printf("%ds:FIGHT AT (%d,%d)/n",second++,state[t].x,state[t].y); second--; } else printf("%ds:(%d,%d)->(%d,%d)/n",second,state[state[t].last].x,state[state[t].last].y,state[t].x,state[t].y);}int main(){ int i,j; while (scanf("%d %d",&n,&m)!=EOF) { for (i=0;i<n;i++) { scanf("%s",g[i]); } init(); //printf("%d",v[0]); bfs(); if (v[n-1][m-1] > 10000) printf("God please help our poor hero./nFINISH/n"); else { printf("It takes %d seconds to reach the target position, let me show you the way./n",v[n-1][m-1]); second = 0; output(flagForEnd); printf("FINISH/n"); } } //getchar(); return 0; }void bfs(){ int i,head = 0,tail = 1; state[head].x = 0; state[head].y =0; state[head].time = 0; state[head].last = -1; v[0][0] = 0; while (head!=tail && tail!=MaxQueue-5) { temp = state[head]; for (i=0;i<4;i++) //search for the 4 direction { int tx = temp.x + dir[i][0]; int ty = temp.y + dir[i][1]; if (tx >= 0 && tx < n && ty >=0 && ty < m && g[tx][ty]!='X') { int plus; if (g[tx][ty] == '.') plus = 1; else plus = g[tx][ty]-'0' + 1; if (temp.time+plus < v[tx][ty]) // 如果有更小步数的走法 { if (tx == n-1 && ty == m-1) flagForEnd = tail; state[tail].x = tx; state[tail].y = ty; v[tx][ty] = temp.time + plus; state[tail].time = v[tx][ty]; state[tail].last = head; tail++; } } } head++; }}void init(){ memset(v,1,sizeof(v));}