UVA 11402 – Ahoy, Pirates!

题目链接

题意：总的来说意思就是给一个01串，然后有3种操作
1、把一个区间变成1
2、把一个区间变成0
3、把一个区间翻转（0变1，1变0）

思路：线段树搞，开一个延迟标记当前操作就可以，注意几种状态间的转变方式就可以

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define lson(x) ((x<<1)+1)#define rson(x) ((x<<1)+2)const int N = 1100005;int t, s[N], n, sn;char str[55];struct Node {    int l, r, b, flag;    Node() {}    Node(int l, int r) {	this->l = l; this->r = r;	b = flag = 0;    }    int len() {	return r - l + 1;    }} node[4 * N];void pushup(int x) {    node[x].b = node[lson(x)].b + node[rson(x)].b;}void build(int l, int r, int x = 0) {    node[x] = Node(l, r);    if (l == r) {	node[x].b = s[l];	return;    }    int mid = (node[x].l + node[x].r) / 2;    build(l, mid, lson(x));    build(mid + 1, r, rson(x));    pushup(x);}void init() {    scanf("%d", &n);    sn = 0;    while (n--) {	int num;	scanf("%d", &num);	scanf("%s", str);	int len = strlen(str);	while (num--) {	    for (int i = 0; i < len; i++)		s[sn++] = str[i] - '0';	}    }    build(0, sn - 1);}//0不变，1黑。2白。3翻转int getS(int a, int b) {    if (a == 3 && b == 3) return 0;    if (a == 3 && b == 1) return 2;    if (a == 3 && b == 2) return 1;    if (a == 3 && b == 0) return 3;    if (a == 2) return 2;    if (a == 1) return 1;    return 0;}void pushdown(int x) {    if (node[x].flag) {	int ls = getS(node[x].flag, node[lson(x)].flag);	node[lson(x)].flag = ls;	if (node[x].flag == 2) node[lson(x)].b = 0;	if (node[x].flag == 1) node[lson(x)].b = node[lson(x)].len();	if (node[x].flag == 3) node[lson(x)].b = node[lson(x)].len() - node[lson(x)].b;	int rs = getS(node[x].flag, node[rson(x)].flag);	node[rson(x)].flag = rs;	if (node[x].flag == 2) node[rson(x)].b = 0;	if (node[x].flag == 1) node[rson(x)].b = node[rson(x)].len();	if (node[x].flag == 3) node[rson(x)].b = node[rson(x)].len() - node[rson(x)].b;	node[x].flag = 0;    }}void set(int l, int r, int v, int x = 0) {    if (node[x].l >= l && node[x].r <= r) {	int ts = getS(v, node[x].flag);	if (v == 1)	    node[x].b = node[x].len();	else if (v == 2)	    node[x].b = 0;	else	    node[x].b = node[x].len() - node[x].b;	node[x].flag = ts;	return;    }    pushdown(x);    int mid = (node[x].l + node[x].r) / 2;    if (l <= mid) set(l, r, v, lson(x));    if (r > mid) set(l, r, v, rson(x));    pushup(x);}int S(int l, int r, int x = 0) {    if (node[x].l >= l && node[x].r <= r)	return node[x].b;    int mid = (node[x].l + node[x].r) / 2;    int ans = 0;    pushdown(x);    if (l <= mid) ans += S(l, r, lson(x));    if (r > mid) ans += S(l, r, rson(x));    pushup(x);    return ans;}int main() {    int cas = 0;    scanf("%d", &t);    while (t--) {	init();	int q;	scanf("%d", &q);	char Q[5]; int a, b;	printf("Case %d:\n", ++cas);	int Qcas = 0;	while (q--) {	    scanf("%s%d%d", Q, &a, &b);	    if (Q[0] == 'F') set(a, b, 1);	    if (Q[0] == 'E') set(a, b, 2);	    if (Q[0] == 'I') set(a, b, 3);	    if (Q[0] == 'S') printf("Q%d: %d\n", ++Qcas, S(a, b));	}    }    return 0;}