[JSOI2010]快递服务

Description

Luogu4046
BZOJ1820

Solution

暴力DP很好想，\(f[i][j][k][l]\)表示处理到第\(i\)个任务，三个人在\(i,j,k\)的方案数。显然一个人应该在这个任务的地方，我们就省去那一维，注意要保证钦定办这个任务的人那一维一定被省略。

Code

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

const int N = 210;
const int INF = 0x3f3f3f3f;

int G[N][N];
int f[2][N][N], n, P[1010];

inline void upd(int &x, const int &y) {
    if (y < x) x = y;
}

int main() {
    n = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j) G[i][j] = read();
    memset(f, 0x3f, sizeof f);
    f[0][2][3] = 0;
    P[0] = 1;
    int nw = 0, pw = 1, x, m = 0;
    while (scanf("%d", &x) != EOF) {
        std::swap(nw, pw);
        P[++m] = x;
        memset(f[nw], 0x3f, sizeof f[nw]);
        for (int j = 1; j <= n; ++j) {
            for (int k = 1; k <= n; ++k) {
                upd(f[nw][j][k], f[pw][j][k] + G[P[m - 1]][x]);
                upd(f[nw][k][P[m - 1]], f[pw][j][k] + G[j][x]);
                upd(f[nw][j][P[m - 1]], f[pw][j][k] + G[k][x]);
            }
        }
    }
    int ans = 0x3f3f3f3f;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            upd(ans, f[nw][i][j]);
        }
    }
    printf("%d\n", ans);
    return 0;
}