HDU-1698-Just a Hook

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ACM模版

描述

题解

典型的线段树问题，区间更新，区间求和，涉及到延时标记。

代码

#include <iostream>using namespace std;const int MAXN = 1e5;struct node{    int left;    int right;    int data;    int sum;};node hook[4 * MAXN];void build(int root, int left, int right){    hook[root].left = left;    hook[root].right = right;    if (left == right)    {        hook[root].data = 1;        hook[root].sum = 1;    }    else    {        int mid = (left + right) >> 1;        build(root * 2, left, mid);        build(root * 2 + 1, mid + 1, right);        hook[root].data = 0;        hook[root].sum = (hook[root * 2].sum + hook[root * 2 + 1].sum);    }}void updata(int root, int left, int right, int data){    if (hook[root].left == left && hook[root].right == right)    {        if (hook[root].data != data)        {            hook[root].data = data;            hook[root].sum = (right - left + 1) * data;        }    }    else    {        if (hook[root].data > 0)        {            hook[root * 2].data = hook[root].data;      //  更新延时标记            hook[root * 2].sum = (hook[root * 2].right - hook[root * 2].left + 1)                                * hook[root * 2].data;            hook[root * 2 + 1].data = hook[root].data;  //  更新延时标记            hook[root * 2 + 1].sum = (hook[root * 2 + 1].right - hook[root * 2 + 1].left + 1)                                * hook[root * 2 + 1].data;        }        hook[root].data = 0;    //  延时标记重置为0        int mid = (hook[root].left + hook[root].right) >> 1;        if (right <= mid)        {            updata(root * 2, left, right, data);        }        else if (left > mid)        {            updata(root * 2 + 1, left, right, data);        }        else        {            updata(root * 2, left, mid, data);            updata(root * 2 + 1, mid + 1, right, data);        }        hook[root].sum = hook[root * 2].sum + hook[root * 2 + 1].sum;    }}int main(){    int T;    scanf("%d",&T);    int N, Q;    int X, Y, Z;    for (int i = 1; i <= T; i++)    {        scanf("%d", &N);        build(1, 1, N);        scanf("%d", &Q);        while (Q--)        {            scanf("%d%d%d", &X, &Y, &Z);            updata(1, X, Y, Z);        }        printf("Case %d: The total value of the hook is %d.\n", i, hook[1].sum);    }    return 0;}

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