iOS MapKit導航及地理轉碼輔助類

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Problem 2167 大王叫我來巡山呐

做的的第二題 呵呵

Problem 2168 防守陣地 I

做的第一題

打了下草稿 馬上切了它

假設當前x=(ai)*1+(ai+1)*2+(ai+2)*3+‘‘‘‘+(aj)*m

下一次是(ai+1)*1+(ai+2)*2+(ai+3)*3+‘‘‘‘+(aj+1)*m = (ai)*1+(ai+1)*2+(ai+2)*3+‘‘‘‘+(aj)*m+(aj+1)*(m+1) -(ai+(ai+1)+(ai+2)+‘‘‘+(aj)+(aj+1))

紫色的這一段是連續的 直接首碼和預先處理

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1000010;int n, m;int a[maxn];int b[maxn];int main(){while(scanf("%d %d", &n, &m) != EOF){__int64 ans = 0;for(int i = 1; i <= n; i++){scanf("%d", &a[i]);b[i] = b[i-1] + a[i];if(i <= m)ans += i*a[i];}__int64 x = ans;for(int i = m+1; i <= n; i++){__int64 temp = x + (m+1) * a[i];temp -= b[i] - b[i-m-1];x = temp;ans = max(ans, temp);}printf("%I64d\n", ans);}return 0;}

 

Problem 2169 shadow

做的第三題

開始想直接廣搜了 不過被消滅的軍隊是不能在被消滅一次的 開始不知道怎麼處理 後來想了下

如果某些軍隊經過最短路的上的點是同一點 那麼接下來他們走的路應該是一樣的 所以直接contine

另外開了個數組sum sum[u]代表到u這個點為止 包括之前的消滅的軍隊總數
 

#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;const int maxn = 100010;int d[maxn];int sum[maxn];int a[maxn];int b[maxn];int vis[maxn];vector <int> G[maxn];int n, m;void BFS(){queue <int> Q;for(int i = 0; i <= n; i++)d[i] = 999999999;memset(sum, 0, sizeof(sum));for(int i = 1; i <= m; i++){Q.push(b[i]);d[b[i]] = 0;vis[b[i]] = 1;}while(!Q.empty()){int u = Q.front(); Q.pop();for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(vis[v]){sum[v] += sum[u];sum[u] = 0;continue;}vis[v] = true;if(d[v] > d[u] + 1){sum[v] += sum[u] + a[v];sum[u] = 0;d[v] = d[u] + 1;Q.push(v);}}}}int main(){while(scanf("%d %d", &n, &m) != EOF){for(int i = 0; i <= n; i++){G[i].clear();}for(int i = 1; i <= n; i++){scanf("%d", &a[i]);}for(int i = 1; i <= m; i++){scanf("%d", &b[i]);}for(int i = 1; i < n; i++){int u, v;scanf("%d %d", &u, &v);G[u].push_back(v);G[v].push_back(u);}memset(vis, 0, sizeof(vis));BFS();int ans = 0;for(int i = 1; i <= n; i++)ans += sum[i];printf("%d\n", ans);}return 0;}

Problem 2170 花生的序列

由於聊聊天 看看動漫沒時間了 本來想看這題的 有空再想想

Problem 2171 防守陣地 II

做的第四題

看了下題目 尼瑪線段樹成段更新！！！！ 直接線段樹搞過去 代碼太噁心 不過真的是水題啊

#include <stdio.h>#define MAX 100010struct node{int l;int r;int sum;int add;}a[MAX*4];void build(int l,int r,int rt){a[rt].l = l;a[rt].r = r;a[rt].add = 0;if(l == r){scanf("%d", &a[rt].sum);return;}int m = (l + r) >> 1;build(l,m,rt<<1);build(m+1,r,rt<<1|1);a[rt].sum = a[rt<<1].sum + a[rt<<1|1].sum;}void update(int l,int r,int rt,int add){if(a[rt].l == l && a[rt].r == r){a[rt].add += add;a[rt].sum += add * (r-l+1);return;}if(a[rt].l == a[rt].r)return;if(a[rt].add){int k = a[rt].r - a[rt].l + 1;a[rt<<1].add += a[rt].add;a[rt<<1|1].add += a[rt].add;a[rt<<1].sum += a[rt].add*(k - (k>>1));a[rt<<1|1].sum += a[rt].add*(k>>1);a[rt].add = 0;}int m = (a[rt].l + a[rt].r) >> 1;if(r <= m)update(l,r,rt<<1,add);else if(l > m)update(l,r,rt<<1|1,add);else{update(l,m,rt<<1,add);update(m+1,r,rt<<1|1,add);}a[rt].sum = a[rt<<1].sum + a[rt<<1|1].sum;}int query(int l, int r, int rt){if(a[rt].l == l && a[rt].r == r){return a[rt].sum;}if(a[rt].l == a[rt].r)return 0;if(a[rt].add){int k = a[rt].r - a[rt].l + 1;a[rt<<1].add += a[rt].add;a[rt<<1|1].add += a[rt].add;a[rt<<1].sum += a[rt].add*(k - (k>>1));a[rt<<1|1].sum += a[rt].add*(k>>1);a[rt].add = 0;}int m = (a[rt].l + a[rt].r) >> 1;int res = 0;if(r <= m)res = query(l,r,rt<<1);else if(l > m)res = query(l,r,rt<<1|1);else{res += query(l,m,rt<<1);res += query(m+1,r,rt<<1|1);}a[rt].sum = a[rt<<1].sum + a[rt<<1|1].sum;return res;}int main(){int n, m, q;while(scanf("%d %d %d",&n, &m, &q) != EOF){build(1, n, 1);while(q--){int x;scanf("%d", &x);printf("%d\n", query(x, x+m-1, 1));update(x, x+m-1, 1, -1);}}return 0;}

 

Problem 2172 巡了南山我巡北山

沒看呢

Problem 2173 Nostop

做的第五題

第一感覺是floyd+矩陣快速冪 10億次floyd肯定不行嘛 矩陣快速冪加速啊

要多做做經典題目啊！！！POJ 3613 這種經典題目很多啊 我們要學經典題目 然後舉一反三

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const __int64 fd = 1;const __int64 INF = fd<<55;const int maxn = 55;struct Mat{    __int64 a[maxn][maxn];};Mat A, B, C, D;int n, m;Mat floyd(Mat x, Mat y){Mat z;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){z.a[i][j] = INF;}for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++) z.a[i][j] = min(z.a[i][j], x.a[i][k] + y.a[k][j]);return z;}/*Mat get(Mat x, Mat y){    Mat z;    memset(z.a, 0, sizeof(z.a));    for(int i = 1; i <= n; i++)        for(int j = 1; j <= n; j++)            for(int k = 1; k <= n; k++)            {                z.a[i][j] += x.a[i][k]*y.a[k][j];                //z.a[i][j] %= mod;            }    return z;}*/void Mat_pow(int x){    while(x)    {        if(x&1)            B = floyd(B, A);        A = floyd(A, A);        x >>= 1;    }}int main(){int T;scanf("%d", &T);while(T--){int x;scanf("%d %d %d", &n, &m, &x);for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){A.a[i][j] = B.a[i][j] = INF;if(i == j)B.a[i][j] = 0;}while(m--){int u, v;__int64 w;scanf("%d %d %I64d", &u, &v, &w);if(A.a[u][v] > w)A.a[u][v] = w;}Mat_pow(x);if(B.a[1][n] == INF)puts("-1");elseprintf("%I64d\n", B.a[1][n]);}return 0;}

 

Problem 2174 卷福的難題

沒看呢

 

 

最後總結一下 前面4題都1A的 最後一題忘記處理-1的情況 坑！！！

不過還是得更加努力 某一方面研究的更深