The Postal Worker Rings Once

Background

Graph algorithms form a very important part of computer science and have a lineage that goes back at least to Euler and the famous Seven Bridges of Königsberg problem. Many optimization problems involve determining efficient methods for reasoning about graphs.

This problem involves determining a route for a postal worker so that all mail is delivered while the postal worker walks a minimal distance, so as to rest weary legs.

The Problem

Given a sequence of streets (connecting given intersections) you are to write a program that determines the minimal cost tour that traverses every street at least once. The tour must begin and end at the same intersection.

The ``real-life'' analogy concerns a postal worker who parks a truck at an intersection and then walks all streets on the postal delivery route (delivering mail) and returns to the truck to continue with the next route.

The cost of traversing a street is a function of the length of the street (there is a cost associated with delivering mail to houses and with walking even if no delivery occurs).

In this problem the number of streets that meet at a given intersection is called the degree of the intersection. There will be at most two intersections with odd degree. All other intersections will have even degree, i.e., an even number of streets meeting at that intersection.

The Input

The input consists of a sequence of one or more postal routes. A route is composed of a sequence of street names (strings), one per line, and is terminated by the string ``deadend'' which is NOT part of the route. The first and last letters of each street name specify the two intersections for that street, the length of the street name indicates the cost of traversing the street. All street names will consist of lowercase alphabetic characters.

For example, the name foo indicates a street with intersections f and o of length 3, and the name computer indicates a street with intersections c and r of length 8. No street name will have the same first and last letter and there will be at most one street directly connecting any two intersections. As specified, the number of intersections with odd degree in a postal route will be at most two. In each postal route there will be a path between all intersections, i.e., the intersections are connected.

The Output

For each postal route the output should consist of the cost of the minimal tour that visits all streets at least once. The minimal tour costs should be output in the order corresponding to the input postal routes.

Sample Input

one
two
three
deadend
mit
dartmouth
linkoping
tasmania
york
emory
cornell
duke
kaunas
hildesheim
concord
arkansas
williams
glasgow
deadend

Sample Output

11
114

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這題較為簡單的中國郵路問題, 因為奇數 degree 的點最多兩個,
因此不需要使用匹配, 只要符合尤拉迴路的條件即可, 讓所有點的 degree 為偶數,
因此對於奇數的點兩兩匹配建立新的邊, 把邊的總長度加上匹配就小就是答案了,
但此題沒那麼複雜, easy !

#include <stdio.h>
#include <string.h>
#define min(x, y) ((x) < (y) ? (x) : (y))
int main() {
    char road[50];
    while(scanf("%s", road) == 1) {
        int map[26][26] = {}, i, j, k;
        int deg[26] = {}, slen, sum = 0;
        memset(map, 63, sizeof(map));
        slen = strlen(road);
        i = road[0]-'a', j = road[slen-1]-'a';
        map[i][j] = min(slen, map[i][j]);
        map[j][i] = min(slen, map[j][i]);
        deg[i]++, deg[j]++;
        sum += slen;
        while(scanf("%s", road) == 1) {
            if(!strcmp("deadend", road))
                break;
            slen = strlen(road);
            i = road[0]-'a', j = road[slen-1]-'a';
            map[i][j] = min(slen, map[i][j]);
            map[j][i] = min(slen, map[j][i]);
            deg[i]++, deg[j]++;
            sum += slen;
        }
        int idx = 0, odeg[2];
        for(i = 0; i < 26; i++)
            if(deg[i]&1)
                odeg[idx++] = i;
        if(idx) {
            for(k = 0; k < 26; k++)
                for(i = 0; i < 26; i++)
                    for(j = 0; j < 26; j++)
                        map[i][j] = min(map[i][j], map[i][k]+map[k][j]);
            sum += map[odeg[0]][odeg[1]];
        }
        printf("%d\n", sum);
    }
    return 0;
}