首页 > ACM题库 > HDU-杭电 > HDU 1208 Pascal’s Travels -动态规划-[解题报告] C++
2013
12-04

HDU 1208 Pascal’s Travels -动态规划-[解题报告] C++

Pascal’s Travels

问题描述 :

An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress.

Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed.


Figure 1


Figure 2

输入:

The input contains data for one to thirty boards, followed by a final line containing only the integer -1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 0-9, with no spaces between them.

输出:

The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 2^63 paths for any board.

样例输入:

4
2331
1213
1231
3110
4
3332
1213
1232
2120
5
11101
01111
11111
11101
11101
-1

样例输出:

3
0
7

Hint
Hint
Brute force methods examining every path will likely exceed the allotted time limit. 64-bit integer values are available as "__int64" values using the Visual C/C++ or "long long" values using GNU C/C++ or "int64" values using Free Pascal compilers.

/*
	map[ i ][ j ]表示跳几格   f [ i ][ j ] 表示有几种条法
	其实就是一个子状态继承问题,如果map[ i ][ j ]为k,
	那么 f [ i+k ][ j ] 和f[i][j+k] 就可以增加 f [ i ][ j ]种跳跃方法了。
*/
#include<stdio.h>
#include<string.h>
char map[55][55];
__int64 f[55][55];
int main()
{
	int n,i,j;
	while(scanf("%d",&n),n!=-1)
	{
		getchar();
		for(i=0;i<n;i++)
			gets(map[i]);//原先没注意数字是连着输入的,按数字读的
		memset(f,0,sizeof(f));
		f[0][0]=1;
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				if(map[i][j]-'0')
				{
					if(i+map[i][j]-'0'<=n)
						f[i+map[i][j]-'0'][j]+=f[i][j];
					if(j+map[i][j]-'0'<=n)
						f[i][j+map[i][j]-'0']+=f[i][j];
				}
		printf("%I64d\n",f[n-1][n-1]);//输出时忘了用%I64d
	}
	return 0;
}


  1. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }