BZOJ 4831 [DP]

Description

给定一个长度为 n 的非负整数序列 a_1,a_2,…a_n 。你可以使用一种操作：选择在序列中连续的两个正整数，并使它们分别减一。当你不能继续操作时游戏结束，而你的得分等于你使用的操作次数。你的任务是计算可能的最小得分和最大得分。

Solution

太强辣，感觉细节比较多。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 1010101;
const int INF = 1 << 30;

inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
inline void read(int &x) {
static char c; x = 0;
for (c = get(); c < '0' || c > '9'; c = get());
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}

int n, test, suf, ans1, ans2;
int a
;
int f
[2][2], g
[2][2];

inline int Min(int a, int b) {
return a < b ? a : b;
}
inline int Max(int a, int b) {
return a > b ? a : b;
}

int main(void) {
read(test);
memset(f, 0x3f, sizeof f);
while (test--) {
read(n);
for (int i = 1; i <= n; i++) read(a[i]);
if (a[1] <= a[2]) {
for (int i = 0; i < a[1]; i++) f[a[2] - i][0][1] = i;
f[a[2] - a[1]][0][0] = a[1];
} else {
for (int i = 0; i <= a[2]; i++) f[a[2] - i][0][1] = i;
}
for (int i = 2; i < n; i++) {
for (int s = 0; s < 2; s++)
for (int j = 0; j <= a[i + 1]; j++)
f[j][i & 1 ^ 1][s] = INF;
for (int s = 0; s < 2; s++) {
for (int j = 0; j <= a[i]; j++) {
if (j <= a[i + 1]) f[a[i + 1] - j][i & 1 ^ 1][0] = Min(f[a[i + 1] - j][i & 1 ^ 1][0], f[j][i & 1][s] + j);
else f[0][i & 1 ^ 1][1] = Min(f[0][i & 1 ^ 1][1], f[j][i & 1][0] + a[i + 1]);
}
if (s == 0) {
suf = INF;
for (int k = a[i]; k > Min(a[i], a[i + 1]); k--) suf = Min(suf, f[k][i & 1][0]);
for (int k = Min(a[i], a[i + 1]); k >= 0; k--) {
f[a[i + 1] - k][i & 1 ^ 1][1] = Min(f[a[i + 1] - k][i & 1 ^ 1][1], suf + k);
suf = Min(suf, f[k][i & 1][0]);
}

}
}
for (int s = 0; s < 2; s++)
for (int j = 0; j <= a[i]; j++)
f[j][i & 1][s] = INF;
}
ans1 = INF;
for (int s = 0; s < 2; s++)
for (int i = 0; i <= a
; i++)
ans1 = Min(ans1, Min(f[0][n & 1][s], f[i][n & 1][0]));
for (int s = 0; s < 2; s++)
for (int i = 0; i <= a
; i++)
f[i][n & 1][s] = INF;
if (a[1] <= a[2]) {
for (int i = 0; i < a[1]; i++) g[a[2] - i][0][1] = i;
g[a[2] - a[1]][0][0] = a[1];
} else {
for (int i = 0; i <= a[2]; i++) g[a[2] - i][0][1] = i;
}
for (int i = 2; i < n; i++) {
for (int s = 0; s < 2; s++)
for (int j = 0; j <= a[i + 1]; j++)
g[j][i & 1 ^ 1][s] = 0;
for (int s = 0; s < 2; s++) {
for (int j = 0; j <= a[i]; j++) {
if (j <= a[i + 1]) g[a[i + 1] - j][i & 1 ^ 1][0] = Max(g[a[i + 1] - j][i & 1 ^ 1][0], g[j][i & 1][s] + j);
else g[0][i & 1 ^ 1][1] = Max(g[0][i & 1 ^ 1][1], g[j][i & 1][0] + a[i + 1]);
}
if (s == 0) {
suf = 0;
for (int k = Min(a[i], a[i + 1]); k >= 0; k--) {
if (k < Min(a[i], a[i + 1])) g[a[i + 1] - k][i & 1 ^ 1][1] = Max(g[a[i + 1] - k][i & 1 ^ 1][1], suf + k);
suf = Max(suf, g[k][i & 1][0]);
}

}
}
for (int s = 0; s < 2; s++)
for (int j = 0; j <= a[i]; j++)
g[j][i & 1][s] = 0;
}
ans2 = 0;
for (int s = 0; s < 2; s++)
for (int i = 0; i <= a
; i++)
ans2 = Max(ans2, Max(g[0][n & 1][s], g[i][n & 1][0]));
for (int s = 0; s < 2; s++)
for (int i = 0; i <= a
; i++)
g[i][n & 1][s] = 0;
printf("%d %d\n", ans1, ans2);
}
return 0;
}