2014 multi-School Union 3 (HDU 4888 HDU 4891 HDU 4893)

HDU 4891 the great pan

Sign-in question: How do you do it? Well, attention. Int will pop up during multiplication.

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;int n;char s[2000000];int flag1, flag2, sp, ansflag;LL ans, tmp;int main() {while (scanf("%d", &n) != EOF) {getchar();ans = 1;tmp = 0;flag1 = 0;flag2 = 0;ansflag = 1;for (int l = 1; l <= n; l++) {gets(s);int len = strlen(s);if (ansflag == 0)continue;for (int i = 0; i < len; i++) {if (flag1 == 0 && flag2 == 0 && s[i] == '{') {flag1 = 1;tmp = 0;} else if (flag1 == 1 && s[i] == '|') {tmp++;} else if (flag1 == 1 && s[i] == '}') {flag1 = 0;ans *= (tmp + 1);tmp = 1;if (ans > 1e5) {ansflag = 0;break;}} else if (flag1 == 0 && flag2 == 0 && s[i] == '$') {flag2 = 1;sp = 0;tmp = 0;} else if (flag2 == 1) {if (s[i] == '$') {if (sp == 1) {ans *= (tmp + 1);if (ans > 1e5) {ansflag = 0;break;}}flag2 = 0;tmp = 0;} else if (sp == 1 && s[i] != ' ') {ans *= (tmp + 1);if (ans > 1e5) {ansflag = 0;break;}sp = 0;tmp = 0;} else if (sp == 0 && s[i] == ' ') {tmp = 1;sp = 1;} else if (sp == 1 && s[i] == ' ') {tmp++;}}}}if (ansflag)printf("%d\n", ans);elseprintf("doge\n");}return 0;}

HDU 4893 wow! Such sequence!
Question:

N numbers are all 0 at the beginning. You have three operations. One operation adds D 2 to the K position, and the other three operations in the range [L, R] output, r] The number of all returns the smallest Fibonacci number closest to it.

Ideas:

Operations 1 and 2 are basically the line segment tree? In order not to time out, the update is obviously delayed. How can we change the value when updating to the [L, R] range?

In fact, the problem can be cleverly solved by storing data

We remember that Val is the vertex value. toval is the sum of the nearest Fibonacci number sum is Val and tosum is toval.

So every three operations, we only need to make the Val and sum of the interval equal to the toval and tosum. During the 1 operation, we also need to update the toval and tosum.

Code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;#define N 100010#define L(x) (x<<1)#define R(x) ((x<<1)|1)#define last 92#define star 1struct node {int l, r, lazy;LL val, toval, sum, tosum;} nd[N * 4];LL F[N];int n, m;LL cal(LL x) {if (x < 0)return -x;return x;}void up(int i) {nd[i].sum = nd[L(i)].sum + nd[R(i)].sum;nd[i].tosum = nd[L(i)].tosum + nd[R(i)].tosum;}void down(int i) {if (nd[i].lazy) {nd[L(i)].val = nd[L(i)].toval;nd[L(i)].sum = nd[L(i)].tosum;nd[L(i)].lazy = 1;nd[R(i)].val = nd[R(i)].toval;nd[R(i)].sum = nd[R(i)].tosum;nd[R(i)].lazy = 1;nd[i].lazy = 0;}}void init(int l, int r, int i) {nd[i].l = l;nd[i].r = r;nd[i].lazy = 0;nd[i].val = 0;nd[i].toval = 0;nd[i].sum = 0;if (l == r) {nd[i].toval = 1;nd[i].tosum = 1;return;}int mid = (l + r) >> 1;init(l, mid, L(i));init(mid + 1, r, R(i));up(i);}void add(int pos, int val, int i) {if (pos == nd[i].l && pos == nd[i].r) {int idx1, idx2;nd[i].val += val;idx1 = lower_bound(F + star, F + last, nd[i].val) - F;idx2 = idx1 - 1;if (cal(F[idx1] - nd[i].val) < cal(F[idx2] - nd[i].val)) {nd[i].toval = nd[i].tosum = F[idx1];nd[i].sum = nd[i].val;} else {nd[i].toval = nd[i].tosum = F[idx2];nd[i].sum = nd[i].val;}nd[i].lazy = 0;return;}down(i);int mid = (nd[i].l + nd[i].r) >> 1;if (pos <= mid)add(pos, val, L(i));elseadd(pos, val, R(i));up(i);}void update(int l, int r, int i) {if (l == nd[i].l && r == nd[i].r) {nd[i].val = nd[i].toval;nd[i].sum = nd[i].tosum;nd[i].lazy = 1;return;}down(i);int mid = (nd[i].l + nd[i].r) >> 1;if (r <= mid)update(l, r, L(i));else if (l > mid)update(l, r, R(i));else {update(l, mid, L(i));update(mid + 1, r, R(i));}up(i);}LL query(int l, int r, int i) {if (l == nd[i].l && r == nd[i].r) {//if(nd[i].lazy) return nd[i].tosum;//else return nd[i].sum;return nd[i].sum;}down(i);int mid = (nd[i].l + nd[i].r) >> 1;LL res = 0;if (r <= mid)res = query(l, r, L(i));else if (l > mid)res = query(l, r, R(i));elseres = query(l, mid, L(i)) + query(mid + 1, r, R(i));if (nd[i].l != nd[i].r)up(i);return res;}void dfs(int i) {printf("l %d r %d lazy %d val %I64d toval %I64d sum %I64d tosum %I64d\n",nd[i].l, nd[i].r, nd[i].lazy, nd[i].val, nd[i].toval, nd[i].sum,nd[i].tosum);if (nd[i].l == nd[i].r)return;dfs(L(i));dfs(R(i));}int main() {int i, op, l, r;F[0] = F[1] = 1;for (i = 2; i < 100; i++)F[i] = F[i - 1] + F[i - 2];//91 7540113804746346429// F [0 ~ 91]while (~scanf("%d%d", &n, &m)) {init(1, n, 1);for (i = 1; i <= m; i++) {scanf("%d%d%d", &op, &l, &r);if (op == 1)add(l, r, 1);else if (op == 2)printf("%I64d\n", query(l, r, 1));elseupdate(l, r, 1);}//dfs(1);}return 0;}

HDU 4888 redraw beautiful drawings

Question:

The number 0 ~ in the N * m Lattice ~ K provides the sum of each row and each column in the lattice (the output value may be multiple or unsolvable in the unique solution)

Ideas:

First of all, we need to think about this question. We can use the network stream to start with the data range or have a good idea.

Next, we create a graph (1) S-> row capacity sum row (2) column-> T capacity sum column (3) row-> column capacity K run Max stream if full stream is at least one solution

Let's take a look at this stream. It means that the traffic passing in from row X goes through the (x, y) lattice and then goes out from column Y. Then, if the full stream is not enough, will it satisfy the question?

Next, determine whether the solution is unique. If the solution is not unique, the following conclusions are available:

Starting from a certain point, at this point + G, at a certain point with it-G, then at a point in the same column of the new point + G, this path is continuously continued like this:

+ G-G

-G + G

Creating a new solution is not like carrying something from a certain point ?? Then move around !! This is the circle, so the solution is not unique !!

In the residual network, if the DFS finds a circle with a length greater than 2, there must be other solutions (move it in this circle !)

DFS is the most violent. This is what teammates say. Basically, four points can be circled. points cannot be too many.

PS: this is a question about dinic time! You can use sap!

If you have ever used dinic, please help me... The dinic I learned from Master Liu should not be written... (Confidence ???

Code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN = 810;const int MAXM = 810 * 810 * 2;const int INF = 0x3f3f3f3f;struct Node {int from, to, next;int cap;} edge[MAXM];int tol;int head[MAXN];int dep[MAXN];int gap[MAXN];int n;void init() {tol = 0;memset(head, -1, sizeof(head));}void addedge(int u, int v, int w) {edge[tol].from = u;edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];head[u] = tol++;edge[tol].from = v;edge[tol].to = u;edge[tol].cap = 0;edge[tol].next = head[v];head[v] = tol++;}void BFS(int start, int end) {memset(dep, -1, sizeof(dep));memset(gap, 0, sizeof(gap));gap[0] = 1;int que[MAXN];int front, rear;front = rear = 0;dep[end] = 0;que[rear++] = end;while (front != rear) {int u = que[front++];if (front == MAXN)front = 0;for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].to;if (dep[v] != -1)continue;que[rear++] = v;if (rear == MAXN)rear = 0;dep[v] = dep[u] + 1;++gap[dep[v]];}}}int SAP(int start, int end) {int res = 0;BFS(start, end);int cur[MAXN];int S[MAXN];int top = 0;memcpy(cur, head, sizeof(head));int u = start;int i;while (dep[start] < n) {if (u == end) {int temp = INF;int inser;for (i = 0; i < top; i++)if (temp > edge[S[i]].cap) {temp = edge[S[i]].cap;inser = i;}for (i = 0; i < top; i++) {edge[S[i]].cap -= temp;edge[S[i] ^ 1].cap += temp;}res += temp;top = inser;u = edge[S[top]].from;}if (u != end && gap[dep[u] - 1] == 0)break;for (i = cur[u]; i != -1; i = edge[i].next)if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1)break;if (i != -1) {cur[u] = i;S[top++] = i;u = edge[i].to;} else {int min = n;for (i = head[u]; i != -1; i = edge[i].next) {if (edge[i].cap == 0)continue;if (min > dep[edge[i].to]) {min = dep[edge[i].to];cur[u] = i;}}--gap[dep[u]];dep[u] = min + 1;++gap[dep[u]];if (u != start)u = edge[S[--top]].from;}}return res;}int vis[MAXN];bool sol(int u, int fa) {int i, v;vis[u] = 1;for (i = head[u]; ~i; i = edge[i].next) {v = edge[i].to;if (!edge[i].cap || v == fa)continue;if (vis[v] || sol(v, u))return true;}vis[u] = 0;return false;}bool findcircle(int fn) {memset(vis, 0, sizeof(vis));for (int i = 1; i <= fn; i++) {vis[i] = 1;if (sol(i, i))return true;vis[i] = 0;}return false;}int sumx[MAXN], sumy[MAXN];int main() {int i, j, k, n1, n2, K, amt, ans;while (~scanf("%d%d%d", &n1, &n2, &K)) {init();sumx[0] = sumy[0] = 0;for (i = 1; i <= n1; i++) {scanf("%d", &sumx[i]);sumx[0] += sumx[i];}for (i = 1; i <= n2; i++) {scanf("%d", &sumy[i]);sumy[0] += sumy[i];}if (sumx[0] != sumy[0]) {puts("Impossible");continue;}for (i = 1; i <= n1; i++) {for (j = 1; j <= n2; j++)addedge(i, n1 + j, K);}amt = tol;for (i = 1; i <= n1; i++)addedge(0, i, sumx[i]);for (i = 1; i <= n2; i++)addedge(n1 + i, n1 + n2 + 1, sumy[i]);n = n1 + n2 + 2;ans = SAP(0, n1 + n2 + 1);if (ans != sumx[0]) {puts("Impossible");continue;}if (findcircle(n1))puts("Not Unique");else {puts("Unique");for (i = 0; i < amt; i += 2) {printf("%d", K - edge[i].cap);if (edge[i].to == n1 + n2)putchar('\n');elseputchar(' ');}}}return 0;}