Codeforces722C-Destroying Array（线段树 or 并查集）

题目链接

http://codeforces.com/contest/722/problem/C

思路

1. 线段树
其实就是线段树的区间并操作，对于每个节点o，需要维护4个值，sumv[o], suml[o], sumr[o], f[o]（分别代表当前区间的最大子区间和，当前区间的左端点到断点的区间和，当前区间的右端点到断点的区间和，以及f代表当前区间是否被分割过）
对于合并操作，见下图：

关键就是pushup()函数
2.并查集
并查集的需要倒跑，先将全部点设置为destroy，然后从最后一个操作开始加点，维护一个最大值
感觉还是线段树的思路更好理解呐= =

代码

1. 线段树

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)const int maxn = 100000 + 5;LL sumv[maxn << 2], suml[maxn << 2], sumr[maxn << 2];bool f[maxn << 2];int pos, n;LL max3(LL x, LL y, LL z) {    return max(x, max(y, z));}void pushup(int o) {    f[o] = f[o << 1] & f[o << 1 | 1];    sumv[o] = max3(suml[o << 1], sumr[o << 1 | 1], sumr[o << 1] + suml[o << 1 | 1]);    sumv[o] = max3(sumv[o], sumv[o << 1], sumv[o << 1 | 1]);    if (f[o << 1]) suml[o] = suml[o << 1] + suml[o << 1 | 1];    else suml[o] = suml[o << 1];    if (f[o << 1 | 1]) sumr[o] = sumr[o << 1 | 1] + sumr[o << 1];    else sumr[o] = sumr[o << 1 | 1];}void build(int o, int L, int R) {    if (L == R) {        scanf("%I64d", &sumv[o]);        suml[o] = sumr[o] = sumv[o];        f[o] = true;        return;    }    int M = (L + R) >> 1;    build(o << 1, L, M);    build(o << 1 | 1, M + 1, R);    pushup(o);}void set(int o, int L, int R) {    if (L == R) {        sumv[o] = suml[o] = sumr[o] = 0;        f[o] = false;        return;    }    int M = (L + R) >> 1;    if (pos <= M) set(o << 1, L, M);    else set(o << 1 | 1, M + 1, R);    pushup(o);}int main() {    scan(n);    build(1, 1, n);    for (int i = 0; i < n; i++) {        scan(pos);        set(1, 1, n);        printf("%I64d\n", max3(sumv[1], suml[1], sumr[1]));    }    return 0;}

1. 并查集

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)const int maxn = 100000 + 5;LL ans[maxn], a[maxn], sumv[maxn];int p[maxn], n, pos[maxn], vis[maxn];int find(int x) {    return p[x] == x ? x : p[x] = find(p[x]);}void merge(int x, int y) {    x = find(x), y = find(y);    if (x != y) {        if (sumv[x] > sumv[y]) {            sumv[x] += sumv[y];            p[y] = x;        } else {            sumv[y] += sumv[x];            p[x] = y;        }    }}int main() {    scan(n);    for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);    for (int i = 1; i <= n; i++) scan(pos[i]);    for (int i = 1; i <= n; i++) p[i] = i;    for (int i = n; i >= 2; i--) {        int pn = pos[i];        vis[pn] = true;        ans[i] = sumv[pn] = a[pn];        if (vis[pn - 1]) {            ans[i] += sumv[find(pn - 1)];            merge(pn, pn - 1);        }        if (vis[pn + 1]) {            ans[i] += sumv[find(pn + 1)];            merge(pn, pn + 1);        }        ans[i] = max(ans[i], ans[i + 1]);    }    for (int i = 2; i <= n; i++) printf("%I64d\n", ans[i]);    printf("0\n");    return 0;}