zoj 2156

/** *  zoj_2156 *  多重背包+遍历路径，优先取小的 */#include <iostream>#include <cstdio>#include <cstring>using namespace std;int const value[] = { 1, 5, 10, 25 };int v[60], n;int coin[4], node[4], p;/* * 应该可以直接用一维代替，要再想想 * 错了，dp[i][coin]应该表示背包容量为coin时，背包中的硬币个数。 *///bool dp[60][10010];int dp[60][10010];bool flag[60];void print_path( int i, int j ) {if ( 0 == i || 0 == j )//return;if ( /*dp[i-1][j] > -1 &&*/ dp[i-1][j] == dp[i][j] ) {      //这里很重要，要不然一直会waprint_path( i-1, j );}else {flag[i] = true;print_path( i-1, j-v[i] );}}int main(){int i, j, k;while ( scanf("%d%d%d%d%d", &p, &coin[0], &coin[1], &coin[2], &coin[3] ) != EOF &&( p+coin[0]+coin[1]+coin[2]+coin[3] != 0 ) ) {n = 0;memset( dp, -1, sizeof(dp) );dp[0][0] = 0;for ( i = 0; i < 4; ++ i ) {k = 1;while ( k < coin[i] ) {v[++n] = value[i] * k;dp[n][0] = 0;coin[i] -= k;for ( j = p; j > 0; -- j ) {dp[n][j] = dp[n-1][j];if ( j >= v[n] && dp[n-1][j-v[n]] > -1 && dp[n][j] < dp[n-1][j-v[n]] + k ) {dp[n][j] = dp[n-1][j-v[n]] + k;}}k <<= 1;}if ( coin[i] > 0 ) {v[++n] = value[i] * coin[i];dp[n][0] = true;for ( j = p; j > 0; -- j ) {dp[n][j] = dp[n-1][j];if ( j >= v[n] && dp[n-1][j-v[n]] > -1 && dp[n][j] < dp[n-1][j-v[n]] + coin[i] ) {dp[n][j] = dp[n-1][j-v[n]] + coin[i];}}}node[i] = n;}//计算路径if ( -1 == dp[n][p] ) {printf("Charlie cannot buy coffee.\n");continue;}memset( flag, 0, sizeof(flag) );print_path( n, p );k = 0;coin[0] = coin[1] = coin[2] = coin[3] = 0;for ( i = 1; i <= node[0]; ++ i ) {if ( false == flag[i] )continue;coin[0] += v[i]/value[0];}for ( j = 1; j < 4; ++ j ) {for ( i = node[j-1]+1; i <= node[j]; ++ i ) {if ( false == flag[i] )continue;coin[j] += v[i]/value[j];}}printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", coin[0], coin[1], coin[2], coin[3] );}return 0;}