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hihoCoder 1251 Today Is a Rainy Day(暴力)

2015-11-16 15:41 489 查看
题目连接:hihoCoder 1251 Today Is a Rainy Day

解题思路

用一个6位6进制表示每个数对应的转换,用广搜预处理代价。

代码

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 50005;
const int inf = 0x3f3f3f3f;

int dp[maxn];

int idx(int* c) {
int ret = 0;
for (int i = 0; i < 6; i++)
ret = ret * 6 + c[i];
return ret;
}

void ridx(int s, int* c) {
for (int i = 5; i >= 0; i--) {
c[i] = s % 6;
s /= 6;
}
}

void presolve() {

int c[10], t[10];
for (int i = 0; i < 6; i++) c[i] = i;
int s = idx(c);

memset(dp, inf, sizeof(dp));
dp[s] = 0;

queue<int> que;
que.push(s);

while (!que.empty()) {
s = que.front();
que.pop();

ridx(s, c);
for (int i = 0; i < 6; i++) {
for (int j = 0; j < 6; j++) {
memcpy(t, c, sizeof(t));
for (int k = 0; k < 6; k++) if (t[k] == i) t[k] = j;
int v = idx(t);
if (dp[v] > dp[s] + 1) {
dp[v] = dp[s] + 1;
que.push(v);
}
}
}
}
}

int G[10][10], C[10];
char a[200], b[200];

int main () {
presolve();
while (scanf("%s%s", a, b) == 2) {
memset(G, 0, sizeof(G));
memset(C, 0, sizeof(C));

int n = strlen(a);
for (int i = 0; i < n; i++) {
int u = b[i] - '1', v = a[i] - '1';
C[u]++;
G[u][v]++;
}

int ans = inf, t[10];
for (int s = 0; s < maxn; s++) {
ridx(s, t);
int tmp = dp[s];
for (int i = 0; i < 6; i++)
tmp += C[i] - G[i][t[i]];
ans = min(ans, tmp);
}
printf("%d\n", ans);
}
return 0;
}
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