题意：给定你海报，覆盖区间，问你最后可以看见多少海报
解题思路：线段树的离散化问题， 把大的数值变成小的数值，然后区间覆盖统计就行了，但是要注意不相邻的数值在离散化的时候要处理，不然会导致 1-10 1-4 6-10 这样结果变为两个海报！
解题代码：
1 #include <stdlib.h> 2 #include <string.h> 3 #include <stdio.h> 4 #define MAXN 51000 5 int hs[MAXN]; 6 struct step 7 { 8 int x, y ; 9 }steps[MAXN]; 10 int qu[MAXN]; 11 struct node 12 { 13 int left ,right ,mid; 14 int cover; 15 }tree[MAXN*6]; 16 struct hst 17 { 18 int num,value; 19 struct hst * next ; 20 }hstp[MAXN]; 21 struct hst *newnode() 22 { 23 struct hst *p; 24 p = (hst*)malloc(sizeof(hst)); 25 p->next = NULL; 26 p->num = 0; 27 p->value = 0 ; 28 return p; 29 } 30 void add(int temp,int value) 31 { 32 int t = temp % 10007; 33 hstp[t].num ++; 34 struct hst *p = newnode(); 35 p->num = temp; 36 p->value = value; 37 p->next = hstp[t].next; 38 hstp[t].next = p ; 39 } 40 int fin(int temp) 41 { 42 int t = temp % 10007; 43 struct hst *p = &hstp[t]; 44 for(int i = 1;i <= hstp[t].num; i ++) 45 { 46 p = p ->next ; 47 if(p->num == temp) 48 return p->value; 49 } 50 51 } 52 53 int cmp(const void *a, const void *b) 54 { 55 return *(int *)a - *(int *)b; 56 } 57 int L(int c) 58 { 59 return c * 2 ; 60 } 61 int R(int c) 62 { 63 return c * 2 + 1; 64 } 65 void build(int c, int p , int v) 66 { 67 tree[c].left = p ; 68 tree[c].right = v; 69 tree[c].mid = (p + v)/2; 70 tree[c].cover = 0 ; 71 if(p == v) 72 return ; 73 build(L(c),p,tree[c].mid); 74 build(R(c),tree[c].mid +1 ,v ); 75 } 76 void update(int c, int p , int v ,int value) 77 { 78 if(p <= tree[c].left && v >= tree[c].right) 79 { 80 tree[c].cover = value; 81 return; 82 } 83 if(tree[c].cover != 0) 84 { 85 tree[L(c)].cover = tree[c].cover; 86 tree[R(c)].cover = tree[c].cover; 87 tree[c].cover = 0 ; 88 } 89 if(v <= tree[c].mid) update(L(c),p,v,value); 90 else if(p > tree[c].mid) update(R(c),p,v,value); 91 else 92 { 93 update(L(c),p,tree[c].mid,value); 94 update(R(c),tree[c].mid+1,v,value ); 95 } 96 } 97 void geths(int c) 98 { 99 if(tree[c].left == tree[c].right || tree[c].cover != 0) 100 { 101 // printf("%d %d %d ",tree[c].left,tree[c].right,tree[c].cover); 102 hs[tree[c].cover] = 1; 103 return ; 104 } 105 geths(L(c)); 106 geths(R(c)); 107 } 108 109 int main() 110 { 111 int t; 112 scanf("%d",&t); 113 while(t--) 114 { 115 memset(hstp,0,sizeof(hstp)); 116 memset(hs,0,sizeof(hs)); 117 memset(qu,0,sizeof(qu)); 118 memset(tree,0,sizeof(tree)); 119 int n; 120 scanf("%d",&n); 121 int qulen = 0 ; 122 for(int i =1 ;i <= n;i ++) 123 { 124 scanf("%d %d",&steps[i].x,&steps[i].y); 125 qulen++; 126 qu[qulen] = steps[i].x; 127 qulen++; 128 qu[qulen] = steps[i].y; 129 } 130 qsort(qu+1,qulen,sizeof(int),cmp); 131 int qutemp = 0; 132 for(int i = 1;i <= qulen;i ++) 133 { 134 if(qu[i] - qu[i-1] == 0) 135 continue; 136 else if(qu[i] - qu[i-1] != 1) 137 { 138 qutemp += 2; 139 add(qu[i],qutemp); 140 } 141 else 142 { 143 qutemp ++; 144 add(qu[i],qutemp); 145 } 146 147 148 } 149 build(1,1,qutemp); 150 for(int i =1 ;i <= n;i ++) 151 { 152 int a,b; 153 a = fin(steps[i].x); 154 b = fin(steps[i].y); 155 // printf("%d %d ",a,b); 156 update(1,a,b,i); 157 } 158 int sum = 0 ; 159 geths(1); 160 for(int i= 1;i <= n;i ++) 161 { 162 if(hs[i] != 0 ) 163 sum ++; 164 } 165 printf("%d ",sum); 166 167 } 168 return 0 ; 169 }
ps：在这题的时候还是没有深入思考离散化的弊端！！