最短路＋拆点 Ａ As long as Binbin loves Sangsang
题意：从１走到ｎ，每次都是LOVE，问到ｎ时路径是连续多个＂LOVE＂的最短距离．秀恩爱不想吐槽．
分析：在普通的最短路上有寻路的限制，把一个点看成４个点，表示通过某一个字符到该点的最短距离．注意自环的处理，还有距离会爆int。
#include <bits/stdc++.h>
const int N = 1314 + 5;
const int M = 13520 + 5;
const long long INF = 200000000000LL;
struct Edge {
int v, w, ch, nex;
};
Edge edge[M<<1];
int head[N];
int n, m, etot;
void init_edge() {
etot = 0;
memset (head, -1, sizeof (head));
}
void add_edge(int u, int v, int w, char ch) {
edge[etot].v = v; edge[etot].w = w;
edge[etot].nex = head[u];
if (ch == 'L') {
edge[etot].ch = 0;
} else if (ch == 'O') {
edge[etot].ch = 1;
} else if (ch == 'V') {
edge[etot].ch = 2;
} else if (ch == 'E') {
edge[etot].ch = 3;
}
head[u] = etot++;
}
struct Node {
int u, id;
};
long long dis[N][4];
int cnt[N][4];
bool vis[N][4];
void SPFA(int s) {
for (int i=1; i<=n; ++i) {
for (int j=0; j<4; ++j) {
vis[i][j] = false;
dis[i][j] = INF;
cnt[i][j] = 0;
}
}
dis[s][3] = 0; vis[s][3] = true;
std::queue<Node> que;
que.push ((Node) {s, 3});
while (!que.empty ()) {
Node now = que.front (); que.pop ();
int u = now.u, id = now.id;
vis[u][id] = false;
for (int i=head[u]; ~i; i=edge[i].nex) {
Edge &e = edge[i];
if (e.ch != (id + 1) % 4) {
continue;
}
if (dis[e.v][e.ch] > dis[u][id] + e.w || dis[e.v][e.ch] == 0) {
dis[e.v][e.ch] = dis[u][id] + e.w;
cnt[e.v][e.ch] = cnt[u][id];
if (e.ch == 3) {
cnt[e.v][e.ch]++;
}
if (!vis[e.v][e.ch]) {
vis[e.v][e.ch] = true;
que.push ((Node) {e.v, e.ch});
}
} else if ((dis[e.v][e.ch] == dis[u][id] + e.w || dis[e.v][e.ch] == 0) && cnt[e.v][e.ch] <= cnt[u][id]) {
cnt[e.v][e.ch] = cnt[u][id];
if (e.ch == 3) {
cnt[e.v][e.ch]++;
}
if (!vis[e.v][e.ch]) {
vis[e.v][e.ch] = true;
que.push ((Node) {e.v, e.ch});
}
}
}
}
}
int main() {
int T;
scanf ("%d", &T);
for (int cas=1; cas<=T; ++cas) {
init_edge ();
scanf ("%d%d", &n, &m);
char ch[2];
for (int i=0; i<m; ++i) {
int u, v, l;
scanf ("%d%d%d%s", &u, &v, &l, ch);
add_edge (u, v, l, ch[0]);
add_edge (v, u, l, ch[0]);
}
SPFA (1);
if (dis[n][3] == INF || cnt[n][3] == 0) {
printf ("Case %d: Binbin you disappoint Sangsang again, damn it!
", cas);
} else {
printf ("Case %d: Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.
", cas, dis[n][3], cnt[n][3]);
}
}
return 0;
}
DP+优化 C Dragon Ball
题意：m个时间，每个时间去取一个龙珠，代价是距离|x-ball[i].pos|，此时人移动到龙珠的位置，还有该龙珠的代价ball[i].cost，问最小代价。
分析：,另一种情况同理，先对每个时间的位置排序，求i-1时间前缀最小的
转移就好了。
#include <bits/stdc++.h>
const int N = 50 + 5;
const int M = 1000 + 5;
const int INF = 0x3f3f3f3f;
struct Ball {
int pos, cost;
bool operator < (const Ball &rhs) const {
return pos < rhs.pos;
}
};
Ball ball[N][M];
int dp[N][M];
int m, n, x;
int main() {
int T;
scanf ("%d", &T);
while (T--) {
scanf ("%d%d%d", &m, &n, &x);
for (int i=1; i<=m; ++i) {
for (int j=1; j<=n; ++j) {
scanf ("%d", &ball[i][j].pos);
}
}
for (int i=1; i<=m; ++i) {
for (int j=1; j<=n; ++j) {
scanf ("%d", &ball[i][j].cost);
}
std::sort (ball[i]+1, ball[i]+1+n);
}
memset (dp, INF, sizeof (dp));
for (int i=1; i<=n; ++i) {
dp[1][i] = abs (ball[1][i].pos - x) + ball[1][i].cost;
}
for (int i=2; i<=m; ++i) {
int k = 1, mn = INF;
for (int j=1; j<=n; ++j) {
for (; k<=n && ball[i-1][k].pos <= ball[i][j].pos; ++k) {
mn = std::min (mn, dp[i-1][k] - ball[i-1][k].pos);
}
dp[i][j] = mn + ball[i][j].pos + ball[i][j].cost;
}
k = n; mn = INF;
for (int j=n; j>=1; --j) {
for (; k>=1 && ball[i-1][k].pos > ball[i][j].pos; --k) {
mn = std::min (mn, dp[i-1][k] + ball[i-1][k].pos);
}
dp[i][j] = std::min (dp[i][j], mn - ball[i][j].pos + ball[i][j].cost);
}
}
int ans = INF;
for (int i=1; i<=n; ++i) {
ans = std::min (ans, dp[m][i]);
}
printf ("%d
", ans);
}
return 0;
}
模拟 E Matrix operation
读懂题，照着模拟
#include <bits/stdc++.h>
int mat[4][4] = {2, 3, 1, 1,
1, 2, 3, 1,
1, 1, 2, 3,
3, 1, 1, 2};
int a[4][4], tmp[4];
void run() {
int ans[4][4];
for (int i=0; i<4; ++i) {
for (int j=0; j<4; ++j) {
for (int k=0; k<4; ++k) {
int t;
if (mat[i][k] == 2) {
t = a[k][j] << 1;
if (t > 0xFF) {
t ^= 0x1B;
}
if (t > 0xFF) {
t %= (0xFF + 1);
}
} else if (mat[i][k] == 3) {
t = a[k][j] << 1;
if (t > 0xFF) {
t ^= 0x1B;
}
t ^= a[k][j];
if (t > 0xFF) {
t %= (0xFF + 1);
}
} else {
t = a[k][j];
}
tmp[k] = t;
}
int t = tmp[0];
for (int i=1; i<4; ++i) {
t ^= tmp[i];
}
ans[i][j] = t;
}
}
for (int i=0; i<4; ++i) {
for (int j=0; j<3; ++j) {
printf ("%02X ", ans[i][j]);
}
printf ("%02X
", ans[i][3]);
}
}
int main() {
int T;
scanf ("%d", &T);
while (T--) {
for (int i=0; i<4; ++i) {
for (int j=0; j<4; ++j) {
scanf ("%X", &a[i][j]);
}
}
run ();
if (T) {
puts ("");
}
}
return 0;
}
数学+快速幂 F Palindrome graph
题意：一个n*n的图填充颜色使得图如何反转或旋转都不会变化。
分析：考虑到是中心对称的，只要考虑1/8的图（三角形）就行了，假设总和x个。填充过的颜色的位置转移到1/8的图中（假设y个），其他的地方能涂k种颜色，答案是k^(x-y)。
#include <bits/stdc++.h>
const int MOD = 100000007;
int pow_mod(int x, int n) {
int ret = 1;
while (n) {
if (n & 1) {
ret = (long long) ret * x % MOD;
}
x = (long long) x * x % MOD;
n >>= 1;
}
return ret;
}
const int N = 10005;
bool vis[N/2][N/2];
int n, m, k;
int main() {
while (scanf ("%d%d%d", &n, &m, &k) == 3) {
memset (vis, false, sizeof (vis));
int mid = n / 2;
if (n & 1) {
mid++;
}
for (int i=0; i<m; ++i) {
int x, y;
scanf ("%d%d", &x, &y);
x++; y++;
if (x > mid) {
x = n + 1 - x;
}
if (y > mid) {
y = n + 1 - y;
}
if (x < y) {
std::swap (x, y);
}
vis[x][y] = true;
}
int c = 0;
for (int i=1; i<=mid; ++i) {
for (int j=1; j<=i; ++j) {
if (!vis[i][j]) {
c++;
}
}
}
printf ("%d
", pow_mod (k, c));
}
return 0;
}
DFS序+线段树 G Successor
题意：给了n个人的上下级关系，每个人有能力和忠诚度，问如果上级被炒了，下属里能力比他强，忠诚度最高的是谁。
分析：其实就是给了一棵树，DFS序转换为线性序列，每一个上级管辖一段区间，按照能力从大到小排序，每次插入能力比上级强的，相同的也同时插入，询问忠诚度最高的对应人的id即是答案。
#include <bits/stdc++.h>
const int N = 5e4 + 5;
struct Person {
int sup, loy, ab, id;
bool operator < (const Person &rhs) const {
return ab > rhs.ab;
}
}p[N];
int n, m;
int left[N], right[N];
std::map<int, int> ID;
std::vector<int> edge[N];
int tot;
#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
int mx[(N<<1)<<2];
void push_up(int o) {
mx[o] = std::max (mx[o<<1], mx[o<<1|1]);
}
void updata(int p, int v, int l, int r, int o) {
if (l == r) {
mx[o] = v;
return ;
}
int mid = l + r >> 1;
if (p <= mid) {
updata (p, v, lson);
} else {
updata (p, v, rson);
}
push_up (o);
}
int query(int ql, int qr, int l, int r, int o) {
if (ql <= l && r <= qr) {
return mx[o];
}
int mid = l + r >> 1, ret = -1;
if (ql <= mid) {
ret = std::max (ret, query (ql, qr, lson));
}
if (qr > mid) {
ret = std::max (ret, query (ql, qr, rson));
}
return ret;
}
void DFS(int u) {
left[u] = tot++;
for (auto v: edge[u]) {
DFS (v);
}
right[u] = tot;
}
int ans[N];
void solve() {
tot = 1;
DFS (0);
//build (1, tot-1, 1);
memset (mx, -1, sizeof (mx));
std::sort (p+1, p+n);
ans[0] = -1;
for (int j, i=1; i<n; i=j) {
j = i;
while (j < n && p[j].ab == p[i].ab) {
int id = p[j].id;
ans[id] = ID[query (left[id]+1, right[id]-1, 1, tot-1, 1)];
j++;
}
j = i;
while (j < n && p[j].ab == p[i].ab) {
int id = p[j].id;
updata (left[id], p[j].loy, 1, tot-1, 1);
j++;
}
}
}
int main() {
int T;
scanf ("%d", &T);
while (T--) {
scanf ("%d%d", &n, &m);
for (int i=0; i<n; ++i) {
edge[i].clear ();
}
ID.clear ();
ID[-1] = -1;
for (int i=1; i<n; ++i) {
scanf ("%d%d%d", &p[i].sup, &p[i].loy, &p[i].ab);
p[i].id = i;
edge[p[i].sup].push_back (i);
ID[p[i].loy] = i;
}
solve ();
for (int i=0; i<m; ++i) {
int j;
scanf ("%d", &j);
printf ("%d
", ans[j]);
}
}
return 0;
}