洛谷P2680 运输计划

大概就是二分+树上差分...

题意:给你树上m条路径,你要把一条边权变为0,使最长的路径最短。

最大的最小,看出二分(事实上我并没有看出来...)

然后二分k,对于所有大于k的边,树上差分求出最长公共边,然后看是否可以。

(yy的解法②:边按照长度排序,然后二分。删除最长公共边。据logeadd juru说是三分)

代码量3.6k,180行,还是有点长的。

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <cstring>
  4 const int N = 300010;
  5 
  6 inline void read(int &x) {
  7     char c = getchar();
  8     x = 0;
  9     while(c > '9' || c < '0') {
 10         c = getchar();
 11     }
 12     while(c <= '9' && c >= '0') {
 13         x = (x << 3) + (x << 1) + c - 48;
 14         c = getchar();
 15     }
 16     return;
 17 }
 18 
 19 struct Edge {
 20     int v, nex, len;
 21 }edge[N << 1]; int top;
 22 
 23 int e[N], n, m, lm, fa[N][20], d[N], lenth[N]; ///
 24 int l[N], r[N], mid[N], len[N]; /// 路径
 25 bool use[N]; /// 树上差分
 26 int num, large, R, f[N];
 27 
 28 inline void add(int x, int y, int z) {
 29     edge[++top].v = y;
 30     edge[top].len = z;
 31     edge[top].nex = e[x];
 32     e[x] = top;
 33     return;
 34 }
 35 
 36 inline void DFS1(int x, int f) {
 37     fa[x][0] = f;
 38     for(int i = e[x]; i; i = edge[i].nex) {
 39         int y = edge[i].v;
 40         if(y != f) {
 41             lenth[y] = lenth[x] + edge[i].len;
 42             d[y] = d[x] + 1;
 43             DFS1(y, x);
 44         }
 45     }
 46     return;
 47 }
 48 
 49 inline void getlca() {
 50     while((1 << lm) < n) {
 51         lm++;
 52     }
 53     DFS1(1, 0);
 54     for(int i = 1; i <= lm; i++) {
 55         for(int x = 1; x <= n; x++) {
 56             fa[x][i] = fa[fa[x][i - 1]][i - 1];
 57         }
 58     }
 59     return;
 60 }
 61 
 62 inline int lca(int x, int y) {
 63     if(d[x] > d[y]) {
 64         std::swap(x, y);
 65     }
 66     int t = lm;
 67     while(t > -1 && d[y] > d[x]) {
 68         if(d[fa[y][t]] >= d[x]) {
 69             y = fa[y][t];
 70         }
 71         t--;
 72     }
 73     if(x == y) {
 74         return x;
 75     }
 76     t = lm;
 77     while(t > -1 && fa[x][0] != fa[y][0]) {
 78         if(fa[x][t] != fa[y][t]) {
 79             x = fa[x][t];
 80             y = fa[y][t];
 81         }
 82         t--;
 83     }
 84     return fa[x][0];
 85 }
 86 
 87 inline int DFS(int x) {
 88     int cnt = 0;
 89     for(int i = e[x]; i; i = edge[i].nex) {
 90         int y = edge[i].v;
 91         if(y == fa[x][0]) {
 92             continue;
 93         }
 94         int temp = DFS(y);
 95         cnt += temp;
 96         if(temp == num) {
 97             large = std::max(large, edge[i].len);
 98         }
 99     }
100     cnt += f[x];
101     return cnt;
102 }
103 
104 inline bool check(int k) {
105     num = 0;
106     memset(f, 0, sizeof(f));
107     for(int i = 1; i <= m; i++) {
108         bool t = len[i] > k;
109         use[i] = t;
110         num += t;
111         if(t) {
112             f[l[i]]++;
113             f[r[i]]++;
114             f[mid[i]] -= 2;
115         }
116     }
117     large = 0;
118     DFS(1);
119     return R - large <= k;
120 }
121 
122 inline int getlong(int i) {
123     int x = l[i];
124     int ans = 0;
125     while(x != mid[i]) {
126         ans = std::max(ans, lenth[x] - lenth[fa[x][0]]);
127         x = fa[x][0];
128     }
129     x = r[i];
130     while(x != mid[i]) {
131         ans = std::max(ans, lenth[x] - lenth[fa[x][0]]);
132         x = fa[x][0];
133     }
134     return ans;
135 }
136 
137 int main() {
138     scanf("%d%d", &n, &m);
139     int x, y, z;
140     for(int i = 1; i < n; i++) {
141         //scanf("%d%d%d", &x, &y, &z);
142         read(x);
143         read(y);
144         read(z);
145         add(x, y, z);
146         add(y, x, z);
147     }
148     getlca();
149     int dr = 0, dl = 0, dm, A = 1;
150     for(int i = 1; i <= m; i++) {
151         //scanf("%d%d", &l[i], &r[i]);
152         read(l[i]);
153         read(r[i]);
154         mid[i] = lca(l[i], r[i]);
155         len[i] = lenth[l[i]] + lenth[r[i]] - 2 * lenth[mid[i]];
156         if(len[i] > dr) {
157             dr = len[i];
158             A = i;
159         }
160     }
161     R = dr;
162     dl = dr - getlong(A);
163     if(dl < 0) {
164         printf("ERROR ");
165     }
166     while(dl < dr) {
167         dm = (dr + dl) / 2;
168         if(check(dm)) {
169             //printf("check %d 1 \n", dm);
170             dr = dm;
171         }
172         else {
173             //printf("check %d 0 \n", dm);
174             dl = dm + 1;
175         }
176     }
177     printf("%d", dr);
178     return 0;
179 }
AC代码

 

posted @ 2018-09-10 21:08  huyufeifei  阅读(129)  评论(0编辑  收藏  举报
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