这个系列总共有7道题，目前只做了3道，gss2比较难，gss4是暴力修改，树状数组维护，还没写，gss6和gss7还不在能力范围内。
SPOJ_1043 GSS1
题意：给定长度不超过5万的序列，M次查询(貌似没给大小？。。)，查询所给区间内的最大子段和。
做法：线段树。维护区间和sum，区间可以得到的最大值smax，从区间最左边开始的最大和lmax和右边开始的rmax，然后就可以了。具体更新还是看代码吧。比较巧妙的在于，把query函数的类型设为线段树的节点类型，这样就可以把它的子区间用update(即pushup)的操作来合并，否则query函数会很难写。
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 int n, m;
5 int a[50100];
6 struct tree{
7 int sum, lmax, rmax, smax;
8 } t[50010 << 2];
9
10 int max(int a, int b)
11 {
12 return a>b?a:b;
13 }
14 void update(tree &x, tree lson, tree rson)
15 {
16 x.sum = lson.sum + rson.sum;
17 x.lmax = max(lson.lmax, lson.sum+rson.lmax);
18 x.rmax = max(rson.rmax, rson.sum+lson.rmax);
19 int tmp = max(x.lmax, x.rmax);
20 tmp = max(tmp, x.sum);
21 tmp = max(tmp, lson.smax);
22 tmp = max(tmp, rson.smax);
23 tmp = max(tmp, lson.rmax+rson.lmax);
24 x.smax = tmp;
25 //x.smax = max(max(x.sum, max(x.lmax, x.rmax)), max(max((lson.smax, rson.smax), (lson.rmax+rson.lmax))));
26 }
27 void build(int x, int l, int r)
28 {
29 if (l == r){
30 t[x].sum = t[x].lmax = t[x].rmax = t[x].smax = a[l];
31 return;
32 }
33 int mid = l + r >> 1, lson = x<<1, rson = x<<1|1;
34 build(lson, l, mid);
35 build(rson, mid+1, r);
36 update(t[x], t[lson], t[rson]);
37 }
38 tree query(int s, int e, int x, int l, int r)
39 {
40 if (s <= l && r <= e){
41 //if (s <= l && r <= e){
42 return t[x];
43 }
44 int mid = l + r >> 1;
45 if (e <= mid) return query(s, e, x<<1, l, mid);
46 if (s > mid) return query(s, e, x<<1|1, mid+1, r);
47 tree p, q, ans;
48 p = query(s, mid, x<<1, l, mid);
49 q = query(mid+1, e, x<<1|1, mid+1, r);
50 update(ans, p, q);
51 return ans;
52 }
53 int main()
54 {
55 scanf("%d", &n);
56 for (int i = 1; i <= n; i++)
57 scanf("%d", &a[i]);
58 build(1, 1, n);
59
60 // for (int i = 1; i <= 10; i++)
61 // printf("%d %d %d %d %d\n", i, t[i].sum, t[i].lmax, t[i].rmax, t[i].smax);
62
63 scanf("%d", &m);
64 int x, y;
65 for (int i = 0; i < m; i++){
66 scanf("%d %d", &x, &y);
67 printf("%d\n", query(x, y, 1, 1, n).smax);
68 }
69 return 0;
70 }
SPOJ_1557 GSS2
SPOJ_1716 GSS3
分析：相比gss1，多了一个单点修改的操作。
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 int n, m;
5 int a[50100];
6 struct tree{
7 int sum, lmax, rmax, smax;
8 } t[50010 << 2];
9
10 int max(int a, int b)
11 {
12 return a>b?a:b;
13 }
14 void pushup(tree &x, tree lson, tree rson)
15 {
16 x.sum = lson.sum + rson.sum;
17 x.lmax = max(lson.lmax, lson.sum+rson.lmax);
18 x.rmax = max(rson.rmax, rson.sum+lson.rmax);
19 /* int tmp = max(x.lmax, x.rmax);
20 tmp = max(tmp, x.sum);
21 tmp = max(tmp, lson.smax);
22 tmp = max(tmp, rson.smax);
23 */
24 int tmp = max(lson.smax, rson.smax);
25 tmp = max(tmp, lson.rmax+rson.lmax);
26 x.smax = tmp;
27 }
28 void build(int x, int l, int r)
29 {
30 if (l == r){
31 t[x].sum = t[x].lmax = t[x].rmax = t[x].smax = a[l];
32 return;
33 }
34 int mid = l + r >> 1, lson = x<<1, rson = x<<1|1;
35 build(lson, l, mid);
36 build(rson, mid+1, r);
37 pushup(t[x], t[lson], t[rson]);
38 }
39 void update(int x, int l, int r, int p, int d)
40 {
41 if (l == r){
42 t[x].sum = t[x].lmax = t[x].rmax = t[x].smax = d;
43 return;
44 }
45 int mid = l + r >> 1;
46 if (p <= mid) update(x<<1, l, mid, p, d);
47 if (p > mid) update(x<<1|1, mid+1, r, p, d);
48 pushup(t[x], t[x<<1], t[x<<1|1]);
49 }
50 tree query(int s, int e, int x, int l, int r)
51 {
52 if (s <= l && r <= e){
53 return t[x];
54 }
55 int mid = l + r >> 1;
56 if (e <= mid) return query(s, e, x<<1, l, mid);
57 if (s > mid) return query(s, e, x<<1|1, mid+1, r);
58 tree p, q, ans;
59 p = query(s, mid, x<<1, l, mid);
60 q = query(mid+1, e, x<<1|1, mid+1, r);
61 pushup(ans, p, q);
62 return ans;
63 }
64 int main()
65 {
66 scanf("%d", &n);
67 for (int i = 1; i <= n; i++)
68 scanf("%d", &a[i]);
69 build(1, 1, n);
70 scanf("%d", &m);
71 int q, x, y;
72 for (int i = 0; i < m; i++){
73 scanf("%d %d %d", &q, &x, &y);
74 if (q == 1)
75 printf("%d\n", query(x, y, 1, 1, n).smax);
76 else
77 update(1, 1, n, x, y);
78 }
79 return 0;
80 }
SPOJ_2713 GSS4
SPOJ_2916 GSS5
分析：相比gss1，查询变成对左端点和右端点的不同区间限制，耐心地分类讨论就可以了。
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 int n, m;
5 int a[10100], sum[10100];
6 struct tree{
7 int sum, lmax, rmax, smax;
8 } t[10100 << 2];
9
10 int max(int a, int b)
11 {
12 return a>b?a:b;
13 }
14 void update(tree &x, tree lson, tree rson)
15 {
16 x.sum = lson.sum + rson.sum;
17 x.lmax = max(lson.lmax, lson.sum+rson.lmax);
18 x.rmax = max(rson.rmax, rson.sum+lson.rmax);
19 int tmp = max(lson.smax, rson.smax);
20 tmp = max(tmp, lson.rmax+rson.lmax);
21 x.smax = tmp;
22 }
23 void build(int x, int l, int r)
24 {
25 if (l == r){
26 t[x].sum = t[x].lmax = t[x].rmax = t[x].smax = a[l];
27 return;
28 }
29 int mid = l + r >> 1, lson = x<<1, rson = x<<1|1;
30 build(lson, l, mid);
31 build(rson, mid+1, r);
32 update(t[x], t[lson], t[rson]);
33 }
34 tree query(int s, int e, int x, int l, int r)
35 {
36 if (s <= l && r <= e){
37 return t[x];
38 }
39 int mid = l + r >> 1;
40 if (e <= mid) return query(s, e, x<<1, l, mid);
41 if (s > mid) return query(s, e, x<<1|1, mid+1, r);
42 tree p, q, ans;
43 p = query(s, mid, x<<1, l, mid);
44 q = query(mid+1, e, x<<1|1, mid+1, r);
45 update(ans, p, q);
46 return ans;
47 }
48 int main()
49 {
50 int T;
51 scanf("%d", &T);
52 while(T--)
53 {
54 scanf("%d", &n);
55 for (int i = 1; i <= n; i++)
56 scanf("%d", &a[i]);
57 sum[0] = 0;
58 for (int i = 1; i <= n; i++)
59 sum[i] = sum[i-1] + a[i];
60 build(1, 1, n);
61 scanf("%d", &m);
62 int x1, y1, x2, y2;
63 for (int i = 0; i < m; i++){
64 scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
65 int ans = 0;
66 if (y1 == x2){
67 // ans = a[y1];
68 // int tmp1 = 0, tmp2 = 0;
69 // if (x1 <= y1-1) tmp1 = max(query(x1, y1-1, 1, 1, n).rmax, 0);
70 // if (x2+1 <= y2) tmp2 = max(query(x2+1, y2, 1, 1, n).lmax, 0);
71 // ans += tmp1 + tmp2;
72 ans = query(x1, y1, 1, 1, n).rmax + query(x2, y2, 1, 1, n).lmax - a[y1];
73 }
74 else if (y1 < x2){
75 ans = sum[x2-1] - sum[y1];
76 ans += query(x1, y1, 1, 1, n).rmax + query(x2, y2, 1, 1, n).lmax;
77 }
78 else{
79 int tmp = query(x2, y1, 1, 1, n).smax;
80 ans = tmp;
81
82 tmp = sum[y1-1] - sum[x2];
83 tmp += query(x1, x2, 1, 1, n).rmax + query(y1, y2, 1, 1, n).lmax;
84 ans = max(tmp, ans);
85
86 tmp = query(x1, x2, 1, 1, n).rmax + query(x2, y1, 1, 1, n).lmax - a[x2];
87 ans = max(tmp, ans);
88
89 tmp = query(x2, y1, 1, 1, n).rmax + query(y1, y2, 1, 1, n).lmax - a[y1];
90 ans = max(tmp, ans);
91
92 }
93 printf("%d\n", ans);
94 }
95 }
96 return 0;
97 }
SPOJ_4487 GSS6
SPOJ_6779 GSS7
SPOJ GSS系列 解题报告,布布扣,bubuko.com
时间： 2024-10-05 16:23:16