soj 2013 weekly-1 7530-7534

7530

题意：填3*3的格子，让给的六个单词在横竖都有。

思路：（AC）深搜生成 P(6,3)的排列，带入检验即可。

#include <iostream>#include <cstring>using namespace std;int arr[6], vis[6], have[6];string given[6], atry[3], tmp;bool flag;void dfs(int cur){if (flag) return ;if (cur == 3){memset(have, 0, sizeof(have));for (int i = 0; i < 3; ++ i){atry[i] = given[arr[i]];have[arr[i]] = 1;}for (int i = 0; i < 3; ++ i){tmp = "";for (int j = 0; j < 3; ++ j)tmp += atry[j][i];for (int j = 0; j < 6; ++ j){if (have[j] == 0 && given[j] == tmp){have[j] = 1;break;}}}bool f = true;for (int i = 0; i < 6; ++ i){if (have[i] == 0){f = false;break;}}if (f){for (int i = 0; i < 3; ++ i)cout << atry[i] << endl;flag = true;}}else{for (int i = 0; i < 6; ++ i) if (!vis[i]){arr[cur] = i;vis[i] = true;dfs(cur+1);vis[i] = false;}}}int main(){while (cin >> given[0]){flag = false;memset(vis, 0, sizeof(vis));for (int i = 1; i < 6; ++ i)cin >> given[i];dfs(0);if (!flag) cout << 0 << endl;}}

7531

题意：n个串，每个串有Ni个环，每个环都可以解开然后套住两个环，问至少需要解开多少个环才能让n个串变成一个串

思路：（AC）每次都在最少的环的串上解开一个环，套住最长的两个串

#include <cstdio>#include <algorithm>using namespace std;int n, arr[500010], top, tail, ans;int main(){while (scanf("%d", &n) != EOF){for (int i = 0; i < n; ++ i)scanf("%d", &arr[i]);sort(arr, arr+n);top = 0;tail = n-1;ans = 0;while (tail > top){arr[top] --;if (arr[top] == 0) top ++;ans ++;tail --;}printf("%d\n", ans);}}

7532

题意：N盘菜，每盘菜都有两个价格Ai，Bi，如果第一个点k菜，那么消费Ai，否则消费Bi，问点1~N盘菜最少需要消费多少（每盘菜做多点一次）

思路：（AC）显然，价格Bi是主价格，我根据加个Bi对所有菜进行排序，对于点K盘菜，有两种可能使消费最少：1. 点前K盘菜，最后消费为前K盘菜的Bi的和 + 前K盘菜中Ai-Bi 最小值 2.消费后N-K盘菜中Ai最小的一盘菜 和 前K-1盘菜。两者取较小值即答案。

#include <vector>#include <cstdio>#include <algorithm>using namespace std;#define N 500000#define Min(a,b) ((a)<(b)?(a):(b))typedef long long LL;typedef pair<int,int> pii;pii data[N];int n, fstMin[N], disMin[N];LL sndSum[N];bool cmp(pii a, pii b){if (a.second < b.second) return true;if (a.second > b.second) return false;return a.first < b.first;}int main(){while (scanf("%d", &n) != EOF){for (int i = 0; i < n; ++ i)scanf("%d%d", &data[i].first, &data[i].second);sort(data, data+n, cmp);sndSum[0] = data[0].second;for (int i = 1; i < n; ++ i)sndSum[i] = sndSum[i-1] + data[i].second;fstMin[n-1] = data[n-1].first;for (int i = n-2; i >= 0; -- i)fstMin[i] = Min( fstMin[i+1], data[i].first );disMin[0] = data[0].first - data[0].second;for (int i = 1; i < n; ++ i)disMin[i] = Min( disMin[i-1], data[i].first-data[i].second );for (int i = 0; i < n; ++ i){long long v1 = sndSum[i] + disMin[i];long long v2 = fstMin[i];if (i != 0) v2 += sndSum[i-1];printf("%lld\n", Min( v1,v2 ));}}}

7533

题意：求1~N这N个数的一个序列，满足M个要求，每个要求描述了位置x~位置y的最大值或者最小值。

思路：（AC）显然这样的序列可能多种。我只要找出每个位置可能的值，再做一次最大匹配即可。

#include <cstdio>#include <cstring>#define N 205#define min(a,b) (a)<(b)?(a):(b)#define max(a,b) (a)<(b)?(b):(a)int n, k, tag, x, y, v;int minn[N], maxn[N], st[N], ed[N];bool bmap[N][N], bmask[N];int cx[N], cy[N];int findpath( int u ) {for (int i = 1; i <= n; ++ i) {if (bmap[u][i] && !bmask[i]) {bmask[i] = true;if (cy[i] == -1 || findpath( cy[i] )) {cy[i] = u;cx[u] = i;return 1;}}}return 0;}bool maxmatch() {int res = 0;memset(cx, -1, sizeof(cx));memset(cy, -1, sizeof(cy));for (int i = 1; i <= n; ++ i) {if (cx[i] == -1) {for (int j = 1; j <= n; ++ j) {bmask[j] = 0;}res += findpath(i);}}return res == n;}int main() {while (scanf("%d%d",&n,&k) != EOF) {for (int i = 1; i <= n; ++ i) {minn[i] = 1;maxn[i] = n;st[i] = 1;ed[i] = n;}while (k --) {scanf("%d%d%d%d",&tag,&x,&y,&v);if (x > y) {int t = x;x = y;y = t;}st[v] = max(st[v], x);ed[v] = min(ed[v], y);if (tag == 1) {for (int i = x; i <= y; ++ i) {maxn[i] = min(maxn[i],v);}}else {for (int i = x; i <= y; ++ i) {minn[i] = max(minn[i],v);}}}memset(bmap, false, sizeof(bmap));for (int i = 1; i <= n; ++ i) {for (int j = minn[i]; j <= maxn[i]; ++ j) {if (st[j] <= i && ed[j] >= i) {bmap[i][j] = true;}}}if (maxmatch()) {for (int i = 1; i <= n; ++ i) {printf("%d%c", cx[i], i==n?'\n':' ');}}else printf("-1\n");}}

7534

题意：有1~N这N个办公楼，给K个指令，指令分两种，1.基础结余为B 日盈利P 的公司在D 天搬入办公楼O，2. 查询X到Y办公楼在D日结余最多的办公楼的结余。当然，如果一个办公楼已经有了公司，但是又搬进了新公司，那么当然结余为新公司的结余，并且新公司要到第二天才开始盈利。查询是在D日办公结束后查询。

思路：（TLE）模拟，每一次询问都更新所有公司的结余。如果是查询操作，就遍历之间所有办公楼找到最大值。

#include <cstdio>typedef long long LL;int n, m;LL maxn;int tag, t, a, b, k, z, s, pt;struct Office {LL balance;int profit;bool flag;void init() { balance = profit = 0; flag = false; }void set(int _b, int _p) { balance = _b; profit = _p; flag = true; }void add(int day) { balance += profit*day; }} office[100005];int main() {freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);while (scanf("%d%d",&n,&m) != EOF) {for (int i = 1; i <= n; ++ i) {office[i].init();}pt = -1;while (m --) {scanf("%d", &tag);if (tag == 1) {scanf("%d%d%d%d",&t,&k,&z,&s);office[k].set(s, z);if (pt) {for (int i = 1; i <= n; ++ i) {if (i != k && office[i].flag) {office[i].add( t-pt );}}}pt = t;}else {scanf("%d%d%d",&t,&a,&b);if (a > b) {tag = a;a = b;b = tag;}if (pt) {for (int i = 1; i <= n; ++ i) {if (office[i].flag) {office[i].add( t-pt );}}}pt = t;bool f = false;for (int i = a; i <= b; ++ i) {if (office[i].flag) {if (!f) {maxn = office[i].balance;f = true;}else if (maxn < office[i].balance) {maxn = office[i].balance;}}}if (f) printf("%lld\n", maxn);else printf("nema\n");}}}}