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POJ 3373 Changing Digits(DP)

2016-10-17 22:12 281 查看
题目链接:点击打开链接

思路:

用d[i][j]表示前i位余数为j的最小修改次数, DP的过程中用path[i][j]表示相同状态下的下一个余数, 目的是记录路径, 用hehe[i][j]表示相同状态下该位最终的值是多少。

---到了现在, AC只为快乐。

细节参见代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 100 + 10;
int T,n,m, d[maxn][10010], vis[maxn][10010], path[maxn][10010], hehe[maxn][10010], kase = 0;
char s[maxn];
int dp(int i, int j) {
int& ans = d[i][j];
if(i == n+1) {
if(j == 0) return 0;
else return INF;
}
if(vis[i][j] == kase) return ans;
vis[i][j] = kase;
ans = INF;
for(int k = 0; k <= 9; k++) {
if(i == 1 && k == 0 && n != 1) continue;
if(k == s[i]-'0') {
int res = (j*10 + s[i]-'0')%m;
int cur = dp(i+1, res);
if(ans > cur) {
ans = cur;
path[i][j] = res;
hehe[i][j] = s[i]-'0';
}
continue;
}
int res = (j*10 + k) % m;
int cur = dp(i+1, res) + 1;
if(ans > cur) {
ans = cur;
path[i][j] = res;
hehe[i][j] = k;
}
}
return ans;
}
void print(int i, int j) {
if(i == n+1) return ;
printf("%d", hehe[i][j]);
print(i+1, path[i][j]);
}
int main() {
while(~scanf("%s%d", s+1, &m)) {
n = strlen(s+1);
++kase;
int ans = dp(1, 0);
print(1, 0);
printf("\n");
}
return 0;
}
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标签:  acm-icpc poj dp