[loj#539][LibreOJ NOIP Round #1]旅游路线_倍增_dp

「LibreOJ NOIP Round #1」旅游路线

题目链接https://loj.ac/problem/539


题解

这个题就很神奇

首先大力$dp$很好想,因为可以把一维放到状态里以取消后效性。

然后就能倍增了...因为就是个智障$dp$

我没想出来/px

代码

#include <bits/stdc++.h>

#define N 110 

#define M 1010 

#define K 30 

#define inf 0x3f3f3f3f 

using namespace std;

int g[N][N][K], p[N], c[N], a[N], b[N], w[N][N], f[N][N * N];

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
    int x = 0, f = 1;
    char c = nc();
    while (c < 48) {
        if (c == '-')
            f = -1;
        c = nc();
    }
    while (c > 47) {
        x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
    }
    return x * f;
}

int main() {
    int n = rd(), m = rd(), C = rd(), T = rd();
    memset(g, 0xef, sizeof g);
    for (int i = 1; i <= n; i ++ ) {
        p[i] = rd(), c[i] = rd();
        c[i] = min(c[i], C);
        g[i][i][0] = 0;
    }
    for (int i = 1; i <= m; i ++ ) {
        int x = rd(), y = rd(), z = rd();
        g[x][y][0] = z;
    }
    for (int k = 1; k <= 17; k ++ ) {
        for (int i = 1; i <= n; i ++ ) {
            for (int j = 1; j <= n; j ++ ) {
                for (int x = 1; x <= n; x ++ ) {
                    g[i][j][k] = max(g[i][j][k], g[i][x][k - 1] + g[x][j][k - 1]);
                }
            }
        }
    }
    for (int i = 1; i <= n; i ++ ) {
        for (int j = 1; j <= n; j ++ ) {
            a[j] = -inf;
        }
        a[i] = 0;
        for (int k = 0; k <= 17; k ++ ) {
            if((c[i] >> k) & 1) {
                for (int j = 1; j <= n; j ++ ) {
                    b[j] = -inf;
                    for (int x = 1; x <= n; x ++ ) {
                        b[j] = max(b[j], a[x] + g[x][j][k]);
                    }
                }
                for (int j = 1; j <= n; j ++ ) {
                    a[j] = b[j];
                }
            }
        }
        for (int j = 1; j <= n; j ++ ) {
            w[i][j] = a[j];
        }
    }
    for (int q = 0; q <= n * n; q ++ ) {
        for (int x = 1; x <= n; x ++ ) {
            if (q >= p[x]) {
                for (int y = 1; y <= n; y ++ ) {
                    f[x][q] = max(f[x][q], f[y][q - p[x]] + w[x][y]);
                }
            }
        }
    }
    while (T -- ) {
        int s = rd(), q = rd(), d = rd();
        int pl = lower_bound(f[s], f[s] + n * n + 1, d) - f[s];
        if (pl > q) {
            puts("-1");
        }
        else {
            printf("%d\n", q - pl);
        }
    }
    return 0;
}

小结:如果是图上的dp,我们可以考虑直接加上一个维度,使得不会出现后效性。

posted @ 2019-08-27 22:53 JZYshuraK_彧 阅读(...) 评论(...) 编辑 收藏