BZOJ 1001 [BeiJing2006]狼抓兔子

[BeiJing2006]狼抓兔子

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