題目鏈接? 2013杭州區域賽
Problem A
Problem B
這題我用的是SPFA+ mask dp
首先跑5次SPFA：
1次是求出每個起點和其他所有點的最短距離
4次是求出每個輸入的點和其他所有點的最短距離
然后就是dp
設dp[mask][i]為，mask狀態下，以i為終點的最優方案
然后做一遍狀壓DP即可。
#include <bits/stdc++.h>using namespace std;#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)const int N = 103;
const int M = 1e4 + 10;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};bool inqueue[M], fl;
int c[N][N], f[6][6], a[M], d[M], dp[101][6];
char s[N][N];
int cnt, n, m, k;
vector <int> v[M];inline void addedge(int x, int y){v[x].push_back(y);v[y].push_back(x);
}void SPFA(int s){memset(inqueue, false, sizeof inqueue);rep(i, 1, cnt) d[i] = 1 << 29;d[s] = 0;queue <int> q;q.push(s);while (!q.empty()){int x = q.front();q.pop();inqueue[x] = false;for (auto u : v[x]){if (d[u] > d[x] + 1){d[u] = d[x] + 1;if (!inqueue[u]){inqueue[u] = true;q.push(u);}}}}
}int main(){while (~scanf("%d%d", &n, &m), n + m){rep(i, 1, n) scanf("%s", s[i] + 1);cnt = 0;rep(i, 1, n) rep(j, 1, m) c[i][j] = ++cnt;scanf("%d", &k);rep(i, 1, n) rep(j, 1, m) if (s[i][j] == '@'){a[k] = c[i][j];s[i][j] = '.';break;}rep(i, 0, k - 1){int x, y;scanf("%d%d", &x, &y);a[i] = c[x][y];}rep(i, 0, n * m + 1) v[i].clear();rep(i, 1, n){rep(j, 1, m) if (s[i][j] == '.'){rep(k, 0, 3){int x = i + dx[k], y = j + dy[k];if (s[x][y] == '.'){addedge(c[i][j], c[x][y]);}}}}SPFA(a[k]);rep(i, 0, k) rep(j, 0, k) f[i][j] = 1 << 29;rep(i, 0, k){SPFA(a[i]);rep(j, 0, k) if (i == j) f[i][j] = 0;else f[i][j] = d[a[j]];}rep(i, 0, 53) rep(j, 0, 5) dp[i][j] = 1 << 29;rep(i, 0, k - 1) dp[1 << i][i] = f[i][k];rep(i, 1, ((1 << k) - 1)){if (__builtin_popcount(i) != 1){rep(j, 0, k - 1) if ((i >> j) & 1){rep(l, 0, k - 1) if (((i >> l) & 1) && (j != l)){dp[i][j] = min(dp[i][j], dp[i ^ (1 << j)][l] + f[j][l]);}}}}int ans = 1 << 30;rep(i, 0, k - 1) ans = min(ans, dp[(1 << k) - 1][i]);printf("%d\n", ans < 1e6 ? ans : -1);}return 0;
}
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Problem C
跟著題意模擬一遍
關鍵是那個旋轉要一次性寫對
#include <bits/stdc++.h>using namespace std;#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)const int N = 41;int a[N][N], b[N][N];
int n;
int ans;void solve(){int c[N][N]; memset(c, 0, sizeof c);rep(i, 1, n) rep(j, 1, n) c[i][j] = a[j][n - i + 1];rep(i, 1, n) rep(j, 1, n) a[i][j] = c[i][j];
}int main(){while (~scanf("%d", &n), n){rep(i, 1, n) rep(j, 1, n) scanf("%d", &a[i][j]);rep(i, 1, n) rep(j, 1, n) scanf("%d", &b[i][j]);ans = 0;rep(p, 1, 4){solve();int now = 0;rep(i, 1, n) rep(j, 1, n) if (a[i][j] == b[i][j]) ++now;ans = max(ans, now);}printf("%d\n", ans);}return 0;
}
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Problem D
Problem E
Problem F
Problem G
Problem H
Problem I
Problem J
Problem K
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轉載于:https://www.cnblogs.com/cxhscst2/p/7624709.html
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