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C. Dasha and Password 预处理 + dp

诗尚凝寒 2022-10-20 阅读 198

​​http://codeforces.com/contest/761/problem/C​​

对于每一个字符串,可以预处理出其到达数字,字母,和特殊符号所需的最小步数。

然后就是在n个东西中,选出数字、字母和特殊符号,至少一个,所需的最少步数。

因为第一个选了数字的话,它就不能选字母的了,然后比赛的时候发现这就是二分图,然后就上了km的模板。(太笨了我)

虽然过了,但是明显正解不是这个。

C. Dasha and Password  预处理 + dp_i++

C. Dasha and Password  预处理 + dp_i++_02

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 500 + 20;
char str[maxn][maxn];
int e[200 + 20][200 + 20];


int match[maxn];//match[col] = row
int vx[maxn], vy[maxn];
int fx[maxn], fy[maxn];
int dfs (int u, int m) {
vx[u] = 1;
int i;
for (i = 1; i <= m; i++) { //筛选n个 导航员 col的值
if (vy[i] == 0 && fx[u] + fy[i] == e[u][i]) {
vy[i] = 1;
if (match[i] == 0 || dfs(match[i], m)) {
match[i] = u; // match[col]=row;
return 1;//搭配成功
}
}
}
return 0;//我找不到啊,后面,就会执行km
}
void do_km(int n, int m) { //
int i, j;
int d = -inf;
for (i = 1; i <= n; i++) { //遍历每一个驾驶员 row的值
if (vx[i] == 1) {
for (j = 1; j <= m; j++) { //对他进行遍历导航员 col的值
if (!vy[j]) {
if (d < (fx[i] + fy[j] - e[i][j])) {
d = fx[i] + fy[j] - e[i][j];
}
}
}
}
}
for (i = 1; i <= n; i++) {
if (vx[i] == 1) {
fx[i] -= d;
vx[i] = 0; //请0
}
if (vy[i] == 1) { //
fy[i] += d;
vy[i] = 0; //情0
}
}
return ;
}
int anskm(int n, int m) {
memset(vx, 0, sizeof(vx));
memset(vy, 0, sizeof(vy));
memset(fx, 0, sizeof(fx));
memset(fy, 0, sizeof(fy));
memset(match, 0, sizeof(match));
//km算法的一部分,先初始化fx,fy
for (int i = 1; i <= n; i++) { //遍历每一个驾驶员 row的值
fy[i] = 0;
fx[i] = inf; //无穷小
for (int j = 1; j <= m; j++) { //遍历每一个导航员 col的值
if (fx[i] > e[i][j]) { //默契值
fx[i] = e[i][j];
}
}
}
for (int i = 1; i <= n; i++) { //遍历每一个驾驶员 row的值
memset(vx, 0, sizeof(vx));
memset(vy, 0, sizeof(vy));
while (!dfs(i, m)) {//如果他找不到搭配,就实现km算法
do_km(n, m);//km完后,还是会对这个想插入的节点进行dfs的,因为他还没搭配嘛
}
}
int ans = 0;
for (int i = 1; i <= m; i++) //遍历导航员,col的值
ans += e[match[i]][i];//输入的row是驾驶员,col是导航员
//match[i]:导航员i和驾驶员match[i]搭配了 match[col]=row;
return ans;
}



void work() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= 200; ++i) {
for (int j = 1; j <= 200; ++j) {
e[i][j] = 500;
}
}
for (int i = 1; i <= n; ++i) {
scanf("%s", str[i] + 1);
int a = 500, b = 500, c = 500;
int len = m;
for (int j = 1; str[i][j]; ++j) {
if (str[i][j] >= '0' && str[i][j] <= '9') {
a = min(a, j - 1);
} else if (str[i][j] == '#' || str[i][j] == '*' || str[i][j] == '&') {
b = min(b, j - 1);
} else {
c = min(c, j - 1);
}
}
for (int j = len; j >= 1; j--) {
if (str[i][j] >= '0' && str[i][j] <= '9') {
a = min(a, len - j + 1);
} else if (str[i][j] == '#' || str[i][j] == '*' || str[i][j] == '&') {
b = min(b, len - j + 1);
} else {
c = min(c, len - j + 1);
}
}
e[i][1] = a;
e[i][2] = b;
e[i][3] = c;
}
// for (int i = 1; i <= n; ++i) {
// for (int j = 1; j <= n; ++j) {
// cout << e[i][j] << " ";
// }
// cout << endl;
// }
cout << anskm(n, n) - (n - 3) * 500 << endl;
}

int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}

km

正解因该是一个dp

dp[i][2][2]][2]表示在前i个字符串中,选出了数字(0 or 1)、字母、特殊符号所需的最小步数。

转移

dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a][b][c]); //保留上次的最小值。

然后对于每一个a、b、c

dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a - 1][b][c] + e[i][1]);之类的。

一开始还担心会不会使得同一个字符串,既选了数字,也选了字母。

然后是不会的,因为需要选数字的时候,是从dp[!now][a - 1][b][c]转移过来的,也就是从上一唯的b、c转过来。所以不会有重复。

C. Dasha and Password  预处理 + dp_i++

C. Dasha and Password  预处理 + dp_i++_02

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 500 + 20;
char str[maxn][maxn];
int e[200 + 20][200 + 20];
int dp[2][2][2][2];
void work() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= 200; ++i) {
for (int j = 1; j <= 200; ++j) {
e[i][j] = 500;
}
}
for (int i = 1; i <= n; ++i) {
scanf("%s", str[i] + 1);
int a = 500, b = 500, c = 500;
int len = m;
for (int j = 1; str[i][j]; ++j) {
if (str[i][j] >= '0' && str[i][j] <= '9') {
a = min(a, j - 1);
} else if (str[i][j] == '#' || str[i][j] == '*' || str[i][j] == '&') {
b = min(b, j - 1);
} else {
c = min(c, j - 1);
}
}
for (int j = len; j >= 1; j--) {
if (str[i][j] >= '0' && str[i][j] <= '9') {
a = min(a, len - j + 1);
} else if (str[i][j] == '#' || str[i][j] == '*' || str[i][j] == '&') {
b = min(b, len - j + 1);
} else {
c = min(c, len - j + 1);
}
}
e[i][1] = a;
e[i][2] = b;
e[i][3] = c;
}
memset(dp, 0x3f, sizeof dp);
int now = 0;
dp[now][0][0][0] = 0;
for (int i = 1; i <= n; ++i) {
now = !now;
for (int a = 0; a <= 1; ++a) {
for (int b = 0; b <= 1; ++b) {
for (int c = 0; c <= 1; ++c) {
dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a][b][c]);
if (a) {
dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a - 1][b][c] + e[i][1]);
// break;
}
if (b) {
dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a][b - 1][c] + e[i][2]);
// break;
}
if (c) {
dp[now][a][b][c] = min(dp[now][a][b][c], dp[!now][a][b][c - 1] + e[i][3]);
// break;
}
}
}
}
}
cout << dp[now][1][1][1] << endl;
}

int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}

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