P1002 过河卒

过河卒

首先初始化地图，去掉不能走的点；
由于卒的走的方向只有向右或向下一步，因此对于点(x,y)，到达该点的路径数为f[x][y] = f[x-1][y] + f[x][y-1]

#include<iostream>#include<cstdio>#include<cstring>long long map[22][22];void init(int x,int y){    if (x>=0&&y>=0) map[x][y] = -1;}int main(){    memset(map,0,sizeof(map));    int a,b;    scanf("%d%d",&a,&b);    int ha,hb;    scanf("%d%d",&ha,&hb);    init(ha,hb);    init(ha+2,hb+1);init(ha+2,hb-1);init(ha+1,hb+2);init(ha+1,hb-2);    init(ha-2,hb+1);init(ha-2,hb-1);init(ha-1,hb+2);init(ha-1,hb-2);    map[0][0] = 1;    for (int i = 0;i <= a;i++){        for (int j = 0;j <= b; j++){            if (i == 0&&j==0) continue;            if (map[i][j] == -1){                map[i][j] = 0;            }else{                if (i-1<0)  map[i][j] = map[i][j-1];                else if (j-1<0) map[i][j] = map[i-1][j];                else map[i][j] = map[i-1][j] + map[i][j-1];            }    //  std::cout<<map[i][j]<<" ";        }        //std::cout<<std::endl;    }    std::cout<<map[a][b];    return 0;}