BZOJ 1001 [BeiJing2006]狼抓兔子

逃离我推掉我的手 2022-01-09 15:02 157阅读 0赞

[\[BeiJing2006\]狼抓兔子][BeiJing2006]

思路：

将边转换成点，每个小三角形两两连边，跑从左下到右上的最短路。

代码：

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    //#define mp make_pair
    #define pb push_back
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdi pair<double, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "\n";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
    
    const int N = 1e3 + 10;
    int a, n, m, cnt = 0;
    int ida[N][N], idb[N][N], idc[N][N], w[N*N*3];
    LL d[N*N*3];
    vector<int> g[N*N*3];
    priority_queue<pli, vector<pli>, greater<pli> > q;
    int main() {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m-1; ++j) scanf("%d", &a), ida[i][j] = ++cnt, w[cnt] = a;
        }
        for (int i = 1; i <= n-1; ++i) {
            for (int j = 1; j <= m; ++j) scanf("%d", &a), idb[i][j] = ++cnt, w[cnt] = a;
        }
        for (int i = 1; i <= n-1; ++i) {
            for (int j = 1; j <= m-1; ++j) scanf("%d", &a), idc[i][j] = ++cnt, w[cnt] = a;
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m-1; ++j) {
                if(i <= n-1) g[ida[i][j]].pb(idb[i][j+1]);
                if(i <= n-1) g[ida[i][j]].pb(idc[i][j]);
                if(i >= 2)   g[ida[i][j]].pb(idb[i-1][j]);
                if(i >= 2)   g[ida[i][j]].pb(idc[i-1][j]);
            }
        }
        for (int i = 1; i <= n-1; ++i) {
            for (int j = 1; j <= m; ++j) {
                if(j <= m-1) g[idb[i][j]].pb(ida[i+1][j]);
                if(j <= m-1) g[idb[i][j]].pb(idc[i][j]);
                if(j >= 1)   g[idb[i][j]].pb(ida[i][j-1]);
                if(j >= 1)   g[idb[i][j]].pb(idc[i][j-1]);
            }
        }
        for (int i = 1; i <= n-1; ++i) {
            for (int j = 1; j <= m-1; ++j) {
                g[idc[i][j]].pb(ida[i][j]);
                g[idc[i][j]].pb(idb[i][j]);
                g[idc[i][j]].pb(ida[i+1][j]);
                g[idc[i][j]].pb(idb[i][j+1]);
            }
        }
        mem(d, 0x3f);
        for (int j = 1; j <= m-1; ++j) {
            d[ida[n][j]] = w[ida[n][j]];
            q.push({w[ida[n][j]], ida[n][j]});
        }
        for (int i = 1; i <= n-1; ++i) {
            d[idb[i][1]] = w[idb[i][1]];
            q.push({w[idb[i][1]], idb[i][1]});
        }
        while(!q.empty()) {
            pli p = q.top();
            q.pop();
            int u = p.se;
            if(d[u] < p.fi) continue;
            for (int i = 0; i < g[u].size(); ++i) {
                int v = g[u][i];
                if(d[v] > d[u] + w[v]) {
                    d[v] = d[u] + w[v];
                    q.push({d[v], v});
                }
            }
        }
        LL ans = 1LL<<60;
        for (int j = 1; j <= m-1; ++j) {
            ans = min(d[ida[1][j]], ans);
        }
        for (int i = 1; i <= n-1; ++i) {
            ans = min(d[idb[i][m]], ans);
        }
        printf("%lld\n", ans);
        return 0;
    }

转载于:https://www.cnblogs.com/widsom/p/10790233.html

[BeiJing2006]: https://www.lydsy.com/JudgeOnline/problem.php?id=1001