BIT1021 Pascal's Travels
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这题本就是个dp水题,但是我做的那个郁闷。。。
用int map[][]居然会TLE,一怒之下换了char map[][],再把getchar什么的换成了cin,就AC了。。。
题意:
4 2331 1213 1231 3110
对这个样例的意思是给一个二维矩阵2 3 3 1
1 2 1 3
1 2 3 1
3 1 1 0
从左上角出发到达右下角的路线有多少条
左上角的2的意思是只能到达距他距离为2的地方,只能往右走和往下走
d[i][j]表示在(i,j)到达终点的路线数
状态转移为d[i][j]=d[i+board[i][j]][j]+d[i][j+board[i][j]]
board是地图
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<stdlib.h>using namespace std;char board[40][40];int n;long long d[40][40];int main(){//freopen("in.txt","r",stdin);while (scanf("%d",&n),n!=-1){for(int i=0;i<n;i++){for (int j = 0; j <n; j++){cin>>board[i][j];}}//输入完成了!!memset(d,-1,sizeof(d));d[n-1][n-1]=1;for (int i=n-1;i>=0;i--){for(int j=n-1;j>=0;j--)//d[i][j]{if(i==n-1&&j==n-1){continue;}d[i][j]=0;//d[i][j]=d[i+board][j]+d[i][j+board]if(i+board[i][j]-'0'>=0&&i+board[i][j]-'0'<n){d[i][j]+=d[i+board[i][j]-'0'][j];}if(j+board[i][j]-'0'>=0&&j+board[i][j]-'0'<n){d[i][j]+=d[i][j+board[i][j]-'0'];}}}printf("%lld\n",d[0][0]);}return 0;}
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