poj-2777

来源：互联网发布：淘宝棉拖鞋女编辑：程序博客网时间：2024/05/21 10:10

// 3756K    719MS   G++ #include <cstdio>#include <cstring>#include <iostream>using namespace std;struct TreeNode {     // 0: no color, >= 1, whole cover by one color, -1: one color    // , but not whole covered, -2: >=1 color, but not whole covered    int color;    int left;    int right;};#define MAX 100010typedef struct TreeNode TreeNode;TreeNode tree[MAX<<2];int L;int opNum;int colorNum;char colorMap[35];void buildTree(int pos, int begin, int end) {    tree[pos].left = begin;    tree[pos].right = end;    tree[pos].color = 1; // initial, all color 1    if (begin == end) {        return;    } else {        int mid = (begin + end)>>1;        buildTree(pos<<1, begin, mid);        buildTree(pos<<1|1, mid+1, end);    }}void insert(int begin, int end, int pos, int color) {    // printf("[%d]\n", tree[2].color);    int curRangeLeft = tree[pos].left;    int curRangeRight = tree[pos].right;    int curColor = tree[pos].color;    // printf("insert %d %d %d %d\n", begin, end, pos, color);    // no cross current range    if (begin > curRangeRight || end < curRangeLeft) {        return;    }    if (begin <= curRangeLeft && end >= curRangeRight) {//whole cover    // printf("insert2 pos: %d %d %d %d\n", pos, curRangeLeft, curRangeRight, color);        tree[pos].color = color;    // printf("2[%d]\n", tree[2].color);    } else { // not whole cover this range        if (curColor == 0) { // first color            tree[pos].color = -1;        } else if (curColor > 0) { // all range is curColor, pushDown            tree[pos<<1].color = curColor;            tree[pos<<1|1].color = curColor;            tree[pos].color = -2; // now has more colors, but no whole covered        } else if (curColor == -1) { // only one color            tree[pos].color = -2;        }        int curRangeMid = (curRangeLeft + curRangeRight)>>1;        if (end <= curRangeMid) { // whole range in left part.            insert(begin, end, pos<<1, color);        } else if (begin <= curRangeMid && end > curRangeMid) { // cross left and right part            insert(begin, curRangeMid, pos<<1, color);            insert(curRangeMid+1, end, pos<<1|1, color);        } else if (begin >= curRangeMid + 1) {            insert(begin, end, pos<<1|1, color);        }    }}int getColorNum(int begin, int end, int pos) {    // printf("3[%d]\n", tree[2].color);    int curRangeLeft = tree[pos].left;    int curRangeRight = tree[pos].right;    int curColor = tree[pos].color;    // printf("getColorNum %d %d %d\n", begin ,end, pos);    // printf("%d %d %d\n", curRangeLeft, curRangeRight, curColor);    if (begin > curRangeRight || end < curRangeLeft) { // no cross        return 0;    } else {        if (curColor > 0) { // only one color cover whole range, return directly            if (!colorMap[curColor]) {                colorMap[curColor] = 1;                return 1;            }        } else if (curColor == 0) { // no color, but has orginal color, return            return 0;        } else { // one or more color, but never cover whole range            int curRangeMid = (curRangeLeft + curRangeRight)>>1;            if (end <= curRangeMid) { // whole range in left part.                return getColorNum(begin, end, pos<<1);            } else if (begin <= curRangeMid && end > curRangeMid) { // cross left and right part                return getColorNum(begin, curRangeMid, pos<<1) +                    getColorNum(curRangeMid+1, end, pos<<1|1);            } else if (begin >= curRangeMid + 1) {                return getColorNum(begin, end, pos<<1|1);            }        }    }    return 0;}int main() {    while(scanf("%d %d %d", &L, &colorNum, &opNum) != EOF) {        // memset(tree, 0, sizeof(tree));        buildTree(1, 1, L);                for (int i = 0; i < opNum; i++) {            // printf("%d %d %d\n", L, colorNum, opNum);            char opType;            // scanf("%c", &opType);            cin>>opType;            if (opType == 'C') {                int begin , end, color;                scanf("%d %d %d", &begin, &end, &color);                // printf("4[%d]\n", tree[2].color);                insert(begin, end, 1, color);                // printf("5[%d]\n", tree[2].color);            } else if (opType == 'P') {                int begin, end;                scanf("%d %d", &begin, &end);                memset(colorMap, 0, sizeof(colorMap));                // printf("6[%d]\n", tree[2].color);                int colorNum = getColorNum(begin, end, 1);                // printf("7[%d]\n", tree[2].color);                printf("%d\n", colorNum);            }        }    }}

线段树经典练手题，经历过2528的洗礼，完全就是水题，思路一模一样，并且连离散化都不用，更简单了。

唯一要注意的就是初始是有颜色的，为1,。

还有我搞了两个状态来标示某个区间有一种/多种颜色，但未完全被某种颜色覆盖的情况，其实完全不用，一种就够了，正两种情况在本题是一样处理逻辑。

这道题也体现线段树的优势，能在大区间完成工作，就不用遍历下面的小区间，大大提高效率。如果是朴素的搞一个数字然后挨个染色，最后挨个检查，其实就有很多的冗余操作.

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