把一个区间内的值全部修改为一个确定值；
计算某个区间内的最大值，最小值，和。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstdlib> 4 using namespace std; 5 #define INF 0x7f7f7f7f 6 #define Max 600000 7 long long a[Max], setv[Max], minv[Max], maxv[Max], sumv[Max], _max, _min, _sum, v; 8 int y1,y2; 9 bool flagv[Max]; 10 inline void maintain(int o, int L, int R) { 11 if (flagv[o]) minv[o] = maxv[o] = setv[o], sumv[o] = (R-L+1)*setv[o]; 12 else if (R > L) minv[o] = min(minv[o*2], minv[o*2+1]), maxv[o] = max(maxv[o*2], maxv[o*2+1]), 13 sumv[o] = sumv[o*2] + sumv[o*2+1]; 14 } 15 void build(int o, int L, int R) { 16 if (L == R) scanf("%lld", a+o), minv[o] = maxv[o] = sumv[o] = setv[o] = a[o]; 17 else {int M = L + (R-L)/2; build(o<<1, L, M), build(o*2+1, M+1, R), maintain(o, L, R);} 18 flagv[o] = false; 19 } 20 inline void pushdown(int o) { 21 if (flagv[o]) setv[o<<1] = setv[o<<1|1] = setv[o], flagv[o<<1] = flagv[o<<1|1] = true, flagv[o] = false; 22 } 23 void update(int o, int L, int R) { 24 if (y1 <= L && y2 >= R) setv[o] = v, flagv[o] = true; 25 else { 26 int M = L + (R-L)/2; pushdown(o); if (y1 <= M) update(o<<1, L, M); else maintain(o<<1, L, M); 27 if (y2 > M) update(o<<1|1, M+1, R); else maintain(o<<1|1, M+1, R); 28 } 29 maintain(o, L, R); 30 } 31 void query(int o, int L, int R) { 32 if (flagv[o]) _sum += setv[o]*(min(y2, R) - max(y1, L) + 1), _min = min(_min, setv[o]), 33 _max = max(_max, setv[o]); 34 else if (y1 <= L && y2 >= R) _sum += sumv[o], _min = min(_min, minv[o]), _max = max(_max, maxv[o]); 35 else {int M = L + (R-L)/2; if (y1 <= M) query(o<<1, L, M); if (y2 > M) query(o<<1|1, M+1, R);} 36 } 37 int main(void) { 38 freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); 39 int n, m; scanf("%d%d", &n, &m); build(1, 1, n); 40 while (m--) { 41 char ch; scanf(" %c", &ch); 42 if (ch == 'Q') scanf("%d%d", &y1, &y2), _sum = 0, _max = -INF, _min = INF, 43 query(1, 1, n), printf("%lld %lld %lld ", _sum, _max, _min); 44 else scanf("%d%d%lld", &y1, &y2, &v), update(1, 1, n); 45 } 46 return 0; 47 }
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