UVa 10066 - The Twin Towers
2014-04-27 22:28
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题目:有两座塔,由不同的石头组成,现在要傻吊一些石头让两个塔一样,求塔最后剩下的最大高度。
分析:dp、最大公共子序列。设状态f(i,j)为串S【0..i-1】与串T【0..j-1】的最大公共子序列。
有状态转移方程: f(i,j)= f(i-1,j-1) {S【i-1】与T【j-1】相同}
f(i,j)= max(f(i-1,j),f(i,j-1)) {S【i-1】与T【j-1】不同}
说明:本来以为是双塔问题,结果是最大公共子序列。
分析:dp、最大公共子序列。设状态f(i,j)为串S【0..i-1】与串T【0..j-1】的最大公共子序列。
有状态转移方程: f(i,j)= f(i-1,j-1) {S【i-1】与T【j-1】相同}
f(i,j)= max(f(i-1,j),f(i,j-1)) {S【i-1】与T【j-1】不同}
说明:本来以为是双塔问题,结果是最大公共子序列。
#include <stdio.h> #include <stdlib.h> #include <string.h> #define max(a,b) ((a)>(b)?(a):(b)) int N1[105],N2[105]; int F[105][105]; int main() { int n1,n2,Case = 1; while ( scanf("%d%d",&n1,&n2) && n1+n2 ) { for ( int i = 1 ; i <= n1 ; ++ i ) scanf("%d",&N1[i]); for ( int i = 1 ; i <= n2 ; ++ i ) scanf("%d",&N2[i]); memset( F, 0, sizeof(F) ); for ( int i = 1 ; i <= n1 ; ++ i ) for ( int j = 1 ; j <= n2 ; ++ j ) if ( N1[i] == N2[j] ) F[i][j] = F[i-1][j-1]+1; else F[i][j] = max( F[i-1][j], F[i][j-1] ); printf("Twin Towers #%d\n",Case ++); printf("Number of Tiles : %d\n\n",F[n1][n2]); } return 0; }
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