codeforces 677D(分层图dp)

Codeforces 677D

传送门：https://codeforces.com/contest/677/problem/D

题意：

给你一个n*m的方格图，每个点有一个权值val，现在要求你从坐标（1，1）开始走，要求你从权值为1的点，走到权值为2的点，依次类推，最终走到权值为p的点的最短路径是多少

题解：

分层图dp

[dp[i][j]表示到达点(i,j)所需要的最短距离是多少\ dis维护一个纵列上的距离\ vis维护一个当前走到的位置\ 转移：dp[r][c] = min(dp[r][c], dis[t][c] + abs(r - t) hspace{1cm} tin[1,n]&&vis[r][c]=i\ dis[r][t] = min(dis[r][t], dp[r][c] + abs(c - t));hspace{1cm} tin[1,m]&&vis[r][t]=i+1\ ]

代码：

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x, y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x, y, z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
struct EDGE {
    int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;

void add_edge(int u, int v) {
    edge[tot].v = v, edge[tot].nxt = head[u], head[u] = tot++;
}

int mp[500][505];
int n, m, k, dp[305][305], vis[305][305], dis[305][305];
vector<pii> b[300 * 300 + 10];

int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, m, p;
    scanf("%d%d%d", &n, &m, &p);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%d", &mp[i][j]);
            b[mp[i][j]].push_back(make_pair(i, j));
        }
    }
    b[0].push_back(make_pair(1, 1));
    memset(dp, 63, sizeof(dp));
    for (int i = 0; i <= p; i++) {
        int len = b[i].size();
        for (int j = 0; j < len; j++) {
            int r = b[i][j].first;
            int c = b[i][j].second;
            if (r == 1 && c == 1 && dp[r][c] == 0) {
                dp[r][c]=INF;
            }
            for (int t = 1; t <= n; t++) {
                if (vis[t][c] == i) {
                    dp[r][c] = min(dp[r][c], dis[t][c] + abs(r - t));
                }
            }
        }
        for (int j = 0; j < len; j++) {
            int r = b[i][j].first;
            int c = b[i][j].second;
            for (int t = 1; t <= m; t++) {
                if (vis[r][t] != i + 1) {
                    vis[r][t] = i + 1;
                    dis[r][t] = dp[r][c] + abs(c - t);
                } else {
                    dis[r][t] = min(dis[r][t], dp[r][c] + abs(c - t));
                }
            }
        }
    }
    printf("%d
", dp[b[p][0].first][b[p][0].second]);

    return 0;
}