ZOJ1610 Count the Colors 经典线段树染色问题

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题意，给你n个 x,y,c，意思就是区间[x,y]被染成C色，但是颜色会被覆盖的，染色操作完成以后问你每种颜色有多少段并输出颜色编号id跟段数cnt

经典问题，不过写的有点撮吧，没去看别人的，这个方法应该是最传统的最普通的，常规的开数组记录，也许大神们有更高端的方法

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#include<cctype>#define ll long long#define LL __int64#define eps 1e-8#define inf 0xfffffff//const LL INF = 1LL<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;const int N = 8000 + 5;typedef struct Node {int l,r;int col;};Node tree[N * 4];int color[N];int ans[N];void init() {memset(color,-1,sizeof(color));memset(ans,0,sizeof(ans));}void build(int l,int r,int id) {tree[id].l = l;tree[id].r = r;tree[id].col = -1;if(l == r)return;int mid = (l + r)/2;build(l,mid,id<<1);build(mid+1,r,id<<1|1);}void cal(int id) {if(tree[id].col > -1) {tree[id<<1].col = tree[id].col;tree[id<<1|1].col = tree[id].col;tree[id].col = -1;}}void update(int l,int r,int id,int col) {if(tree[id].col == col)return;if(l <= tree[id].l && r >= tree[id].r) { tree[id].col = col;return;}cal(id);int mid = (tree[id].l + tree[id].r)/2;if(r <= mid)update(l,r,id<<1,col);else if(l > mid)update(l,r,id<<1|1,col);else {update(l,mid,id<<1,col);update(mid+1,r,id<<1|1,col);}}void query(int l,int r,int id) {if(tree[id].col > -1) {for(int i=tree[id].l;i<=tree[id].r;i++)color[i] = tree[id].col;return;}if(tree[id].l == tree[id].r)return;int mid = (tree[id].l + tree[id].r)/2;if(r <= mid)query(l,r,id<<1);else if(l > mid)query(l,r,id<<1|1);else {query(l,mid,id<<1);query(mid+1,r,id<<1|1);}}int main() {int n;while(scanf("%d",&n) == 1) {int q = n;init();build(1,N,1);int x,y,c;while(q--) {scanf("%d %d %d",&x,&y,&c);x++;update(x,y,1,c);}query(1,N,1);for(int i=1;i<N;i++)if(color[i - 1] != color[i])if(color[i] > -1)ans[color[i]]++;for(int i=0;i<N;i++)if(ans[i])printf("%d %d\n",i,ans[i]);puts("");}return 0;}

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