UVA 1379 - Pitcher Rotation(DP + 贪心)
题目链接:1379 - Pitcher Rotation
题意:n个人,m个敌人。去比赛,有得分。n个人能够反复比。可是每次比完要歇息4天,问最大得分
思路:dp[i][j][k][l][x] 表示第场比赛,前一天为j,两天为k,三天为l。四天为x,的最大得分,然后因为仅仅有每一个人5天就能用一次,所以对于每一个人来说,仅仅有得分前5的会被使用上。所以后4维状态仅仅须要5^4,进行状态转移。不用比赛的情况分开考虑。还有这题内存有限,要用滚动数组优化不然会RE
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
const int N = 105;
const int M = 205;
int t, n, m, G, d[M + 10], i, x, j, k, l, y, dp[2][6][6][6][6], ans;
struct Map {
int v, id;
} g[N][N];
bool cmp(Map a, Map b) {
return a.v > b.v;
}
int cal(int u, int v) {
if (u == 0 || v == 0)
return 0;
return g[u][v].id;
}
int main() {
scanf("%d", &t);
while (t--) {
ans = 0;
scanf("%d%d%d", &n, &m, &G);
for (i = 1; i <= m; i++) {
for (j = 1; j <= n; j++) {
scanf("%d", &g[i][j].v);
g[i][j].id = j;
}
sort(g[i] + 1, g[i] + 1 + n, cmp);
}
G += 10;
for (i = 1; i <= G; i++)
scanf("%d", &d[i]);
memset(dp[0], 0, sizeof(dp[0]));
for (i = 1; i <= G; i++) {
int now = i % 2;
int pre = (!now);
memset(dp[now], 0, sizeof(dp[now]));
if (d[i]) {
for (y = 1; y <= 5; y++) {
for (j = 0; j <= 5; j++) {
if (i > 1 && cal(d[i], y) == cal(d[i - 1], j)) continue;
for (k = 0; k <= 5; k++) {
if (i > 2 && cal(d[i], y) == cal(d[i - 2], k)) continue;
for (l = 0; l <= 5; l++) {
if (i > 3 && cal(d[i], y) == cal(d[i - 3], l)) continue;
for (x = 0; x <= 5; x++) {
if (i > 4 && cal(d[i], y) == cal(d[i - 4], x)) continue;
dp[now][y][j][k][l] = max(dp[now][y][j][k][l], dp[pre][j][k][l][x] + g[d[i]][y].v);
ans = max(ans, dp[now][y][j][k][l]);
}
}
}
}
}
}
else {
for (j = 0; j <= 5; j++) {
for (k = 0; k <= 5; k++) {
for (l = 0; l <= 5; l++) {
for (x = 0; x <= 5; x++) {
dp[now][0][j][k][l] = max(dp[now][0][j][k][l], dp[pre][j][k][l][x]);
ans = max(ans, dp[now][0][j][k][l]);
}
}
}
}
}
}
printf("%.2lf\n", ans * 1.0 / 100);
}
return 0;
}